ML Aggarwal Solution Class 10 Chapter 9 Arithmetic and Geometric Progressions Test

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 Q31 Q32 Q33 Q34 Q35 Q36 Q37

 Test

Question 1

Write the first four terms of the A.P. when its first term is – 5 and the common difference is – 3.

Sol :

First 4 term of A.P. whose first term (a) = -5

and common difference (d) = -3

= -5, -8, -11, -14

Question 2

Verify that each of the following lists of numbers is an A.P., and the write its next three terms :

(i) $0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots$

(ii) $5, \frac{14}{3}, \frac{13}{3}, 4, \ldots$

Sol :

(i) $0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots$

Here $a=0, d=\frac{1}{4}$

$\therefore$ Next three terms will be $1, \frac{5}{4}, \frac{3}{2}$

(ii) $5, \frac{14}{3}, \frac{13}{3}, 4, \ldots$

Here, $a=5, d=\frac{14}{3}-5=\frac{14-15}{3}=\frac{-1}{3}$

$\therefore$ Next three terms will be

$a_{2}=4-\frac{1}{3}=\frac{11}{3}$

$a_{3}=\frac{11}{3}-\frac{1}{3}=\frac{10}{3}$

$a_{4}=\frac{10}{3}-\frac{1}{3}=\frac{9}{3}=3$

i.e. $\frac{11}{3}, \frac{10}{3}, 3$

Question 3

The nth term of an A.P. is 6n + 2. Find the common difference.

Sol :

Tn of an A.P. = 6n + 2 .

$T_1$ = 6 x 1 + 2 = 6 + 2 = 8

$T_2$ = 6 x 2 + 2 = 12 + 2 = 14

$T_3 $= 6 x 3 + 2 = 18 + 2 = 20

d = 14 – 8 = 6

Question 4

Show that the list of numbers 9, 12, 15, 18, … form an A.P. Find its 16th term and the nth.

Sol :

9, 12, 15, 18, …

Here, a = 9, d = 12 – 9 = 3

or 15 – 12 = 3

or 18 – 15 = 3

Yes, it form an A.P.

$\mathrm{T}_{16}=a+(n-1) d=9+(16-1) \times 3$

$=9+15 \times 3=9+45=54$

and $\mathrm{T}_{n}=a+(n-1) d=9+(n-1) \times 3$

$=9+3 n-3=3 n+6$

Question 5

Find the 6th term from the end of the A.P. 17, 14, 11, …, – 40.

Sol :

6th term from the end of

A.P. = 17, 14, 11, …… 40

Here, a = 17, d = -3, l = -40

l = a + (n – 1 )d

l=a+(n-1) d

-40=17+(n-1)(-3)

-40=17+(n-1)(-3)

-40-17=(n-1)(-3)

$\frac{-57}{-3}=n-1$

19=n-1

n=19+1=20

$\therefore$ 6th term from the end

=l-(n-1) d

=-40-(6-1)(-3)

=-40+15=-25

Question 6

If the 8th term of an A.P. is 31 and the 15th term is 16 more than its 11th term, then find the A.P.

Sol :

In an A.P.

$a_{8}=31, a_{15}=a_{11}+16$

Let a be the first term and d be a common difference, then

$a_{8}=a+(n-1) d=31 \Rightarrow a+7 d=31 \ldots(i)$

Similarly,

$a_{15}=a+14 d=a+10 d+16$

14d-10d=16 

$\Rightarrow 4 d=16$

$\Rightarrow d=\frac{16}{4}=4$

From

(i) $a+7 \times 4=31$

$\Rightarrow a+28=31 \Rightarrow a=31-28=3$

$\therefore a=3, d=4$

Now, A.P. will be $3,7,11,15, \ldots$

Question 7

The 17th term of anA.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, then find the wth term.

Sol :

In an A.P.

$a_{17}=2 \times a_{8}+5$

$a_{11}=43,$ find $a_{n}$

Let a be the first term and d be the common

difference, then

$a_{11}$=a+(n-1)d

=a+(11-1)d

=a+10d=43..(i)

Similarly,

$a_{17}=2 \times a_{8}+5$

a+16d=2(a+7d)+5

a+16d=2a+14d+5

-5+16a-14d=2a-a 

$\Rightarrow a=2 d-5$..(ii)

From (i) and (ii)

2d-5+10d=43 

$\Rightarrow 12 d=43+5=48$

$d=\frac{48}{12}=4$

But a+10d=43

$\therefore a+10 \times 4=43 \Rightarrow a+40=43$

$\Rightarrow a=43-40=3$

$\therefore a=3, d=4$

Now, $a_{n}=a+(n-1) d$

=3+4(n-1)

=3+4n-4

=4n-1

Question 8

The 19th term of an A.P. is equal to three times its 6th term. If its 9th term is 19, find the A.P.

Sol :

In an A.P.

$a_{19}=3 \times a_{6}$ and $a_{9}=19$

Let a be the first term and d be the common

difference, then

$a_{9}=a+(n-1) d=a+(9-1) d=a+8 d$

a+8d=19..(i)

Similarly,

$a_{19}=3 \times a_{6}$

$\Rightarrow a+18 d=3(a+5 d)$

$a+18 d=3 a+15 d $

$\Rightarrow 3 a-a=18 d-15 d$

$\Rightarrow 2 a=3 d$..(ii)

$a=\frac{3}{2} d$

From (i)

$\frac{3}{2} d+8 d=19 $

$\Rightarrow \frac{19}{2} d=19$

$\Rightarrow d=\frac{19 \times 2}{19}=2$

and $a=\frac{3}{2} d=\frac{3}{2} \times 2=3$

$\therefore a=3, d=2$ and A.P. is $3,5,7,9, \ldots$

Question 9

If the 3rd and the 9th terms of an A.P. are 4 and – 8 respectively, then which term of this A.P. is zero?

Sol :

In an A.P.

$a_{3}=4, a_{9}=-8$, which term of A.P. will be zero

Let a be the first term and d be a common difference, then

$a_{3}=a+(n-1) d=a+(3-1) d$

$\Rightarrow a+2 d=4$

Similarly, a+8d=-8

Subtracting, we get

6d=-12

$ \Rightarrow d=\frac{-12}{6}=-2$

and a+2d=4 

$\Rightarrow a+2 \times(-2)=4$

$\Rightarrow a-4=4$

$ \Rightarrow a=4+4=8$

Let n th term be zero, then

a+(n-1) d=0

$ \Rightarrow 8+(n-1) \times(-2)=0$

$\Rightarrow-2 n+2=-8$

$ \Rightarrow-2 n=-8-2=-10$

$\Rightarrow n=\frac{-10}{-2}=5$

$\therefore 0$ will be the fifth term.

Question 10

Which term of the list of numbers 5, 2, – 1, – 4, … is – 55?

Sol :

A.P. is 5, 2, -1, – 4, …

Which term of A.P. is -55

Let it be nth term

Here, a = 5, d = 2 – 5 = -3

$\therefore a_{n}=a+(n-1) d$

$\Rightarrow-55=5+(n-1) \times(-3)$

$-55-5=-3(n-1) $

$\Rightarrow \frac{-60}{-3}=n-1$

$\Rightarrow n-1=20$

$ \Rightarrow n=20+1=21$

$\therefore-55$ is the 21 st term.

Question 11

The 24th term of an A.P. is twice its 10th term. Show that its 72nd term is four times its 15th term.

Sol :

In an A.P.

24th term = 2 x 10th term

To show that 72nd term = 4 x 15th term

Let a be the first term and d be a common difference, then

24 th term =a+(24-1)d

=a+23d

and 10 th term =a+9d

$\therefore a+23 d=2(a+9 d)$

$\Rightarrow a+23 d=2 a+18 d$

$\Rightarrow 2 a-a=23 d-18 d$

$\Rightarrow a=5 d$...(i)

and 72 nd term =a+71d

and 15 th term =a+14d

Substitute the value of (i), we get

a+71 d=5 d+71 d=76 d

and a+14 d=5 d+14 d=19 d

$\therefore 76 d=4 \times 19 d$

Hence 72 nd term is 4 times the 15 th term.

Question 12

Which term of the list of numbers $20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots$  is the first negative term?

Sol :

In A.P., which is the first negative term

$20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots$

Here, $a=20, d=19 \frac{1}{4}-20=\frac{-3}{4}$

Let n th term be first negative term

$\therefore a_{n}=a+(n-1) d$

Let n th term be first negative term, then

$a_{n}=20+(n-1)\left(\frac{-3}{4}\right)$

$\Rightarrow a_{n}=20+(n-1)\left(\frac{-3}{4}\right)$

$\Rightarrow a_{n}=20-\frac{3}{4} n+\frac{3}{4}$

Now, $a_{n}<0$ is the first negative term

$\Rightarrow 20+\frac{3}{4}-\frac{3}{4} n<0 $

$\Rightarrow \frac{83}{4}-\frac{3}{4} n<0$

$\Rightarrow \frac{83}{4}<\frac{3}{4} n \Rightarrow 83<3 n$

$\Rightarrow \frac{83}{3}<n \Rightarrow 28<n$

$\therefore 28$ th is the first negative term.

Question 13

If the pth term of an A.P. is q and the qth term is p, show that its nth term is (p + q – n)

Sol :

In an A.P.

pth term = q

qth term = p

Show that (p + q – n) is nth term

Let a be the first term and d be the common

difference

$\therefore p$ th term =a+(p-1)d=q..(i)

and q th term =a+(q-1)d=p..(ii)

Subtracting, we get

q-p=(p-1-q+1)d=(p-q)d

$d=\frac{q-p}{p-q}=\frac{-(p-q)}{(p-q)}=-1$

From (i), $a+(p-1) \times(-1)=q$

a-p+1=q

a=q+p-1

L.H.S

nth term

=a+(n-1) d=(p+q-1)+(n-1)(-1)

=p+q-1-n+1=p+q-n=R.H.S

Question 14

How many three digit numbers are divisible by 9?

Sol :

3-digit numbers which are divisible by 9 are 108, 117, 126, 135, …, 999

Here, a = 108, d = 9 and l = 999

$\therefore l=a_{n}=a+(n-1) d$

$\Rightarrow 999=108+(n-1) 9$

$\Rightarrow 999-108=9(n-1)$

$\Rightarrow 891=9(n-1) $

$\Rightarrow \frac{891}{9}=n-1$

$\Rightarrow n-1=99 $

$\Rightarrow n=99+1=100$

$\therefore$ There are 100 numbers or terms.

Question 15

The sum of three numbers in A.P. is – 3 and the product is 8. Find the numbers.

Sol :

Sum of three numbers of an A.P. = -3

and their product = 8

Let the numbers be

a – d, a, a + d, then

a – d + a + a + d = -3

$\Rightarrow 3 a=-3 \Rightarrow a=\frac{-3}{3}=-1$

and (a-d) a(a+d)=8

Question 16

The angles of a quadrilateral are in A.P. If the greatest angle is double of the smallest angle, find all the four angles.

Sol :

Angles of a quadrilateral are in A.P.

Greatest angle is double of the smallest

Let the smallest angle of the quadrilateral is

a+3d…..(i)

and other are a+d, a-d, a-3d

Where a-3d is the smallest

$\therefore a+3 d=2(a-3 d)$

$\Rightarrow a+3 d=2 a-6 d$

2a-a=3d+6d

$\Rightarrow a=9 d$..(ii)

But sum of angles of a quadrilateral =360°

$\therefore a-3 d+a-d+a+d+a+3 d=360^{\circ}$

$4 a=360^{\circ} \Rightarrow a=\frac{360^{\circ}}{4}=90^{\circ}$

$\therefore 9 d=a=90^{\circ} \Rightarrow d=\frac{90^{\circ}}{9}=10^{\circ}$ [From (ii)]

Greatest angle $=a+3 d=90^{\circ}+30^{\circ}=120^{\circ}$

Other angles are $=a+d=90^{\circ}+10^{\circ}=100^{\circ}$

$a-d=90^{\circ}-10^{\circ}=80^{\circ}$

and $a-3 d=90^{\circ}-30^{\circ}=60^{\circ}$

Hence angles are $60^{\circ}, 80^{\circ}, 100^{\circ}, 120^{\circ}$

Question 17

The nth term of an A.P. cannot be n² + n + 1. Justify your answer.

Sol :

nth term of an A.P. can’t be n² + n + 1

Giving some different values to n such as 1, 2, 3, 4, …

we find then

$a_{1}=1^{2}+1+1=1+1+1=3$

$a_{2}=2^{2}+2+1=4+2+1=7$

$a_{3}=3^{2}+3+1=9+3+1=13$

$a_{4}=4^{2}+4+1=16+4+1=21$

We see that,

$d=a_{2}-a_{1}=7-3=4$

$d=a_{3}-a_{2}=13-7=6$

$d=a_{4}-a_{3}=21-13=8$

We see that d is not constant

$\therefore$ It is not an A.P.

Hence, $a_{n} \neq n^{2}+n+1$

Question 18

Find the sum of first 20 terms of an A.P. whose nth term is 15 – 4n.

Sol :

Giving some different values such as 1 to 20

We get,

$a_{1}=15-4 \times 1=15-4=11$

$a_{2}=15-4 \times 2=15-8=7$

$a_{3}=15-4 \times 3=15-12=3$

$a_{4}=15-4 \times 4=15-16=-1$

and so on,

$a_{20}=15-4 \times 20=15-80=-65$

Now, A.P. is $11,7,3,-1, \ldots,-65$

Here, a=11, d=-4 and n=20

$\mathrm{S}_{20}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{20}{2}[2 \times 11+(20-1)(-4)]$

=10[22-76]

=10(-54)=-540

Question 19

Find the sum :

$18+15 \frac{1}{2}+13+\ldots+\left(-49 \frac{1}{2}\right)$

Sol :

$18+15 \frac{1}{2}+13+\ldots+\left(-49 \frac{1}{2}\right)$

Here, $a=18, d=15 \frac{1}{2}-18=-2 \frac{1}{2}=\frac{-5}{2}$

$l=-49 \frac{1}{2}=\frac{-99}{2}$

$a_{n}=a+(n-1) d$

$\frac{-99}{2}=18+(n-1)\left(\frac{-5}{2}\right)$

$\frac{-99}{2}-\frac{18}{1}=\frac{-5}{2}(n-1)$

$\frac{-99-36}{2}=\frac{-5}{2}(n-1)$

$\Rightarrow \frac{-135}{2}=\frac{-5}{2}(n-1)$

$\Rightarrow \frac{-135}{2} \times \frac{2}{-5}=n-1 $

$\Rightarrow n-1=27$

$\Rightarrow n=27+1=28$

Now, $\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\mathrm{S}_{28}=\frac{28}{2}\left[2 \times 18+(28-1)\left(\frac{-5}{2}\right)\right]$

$\mathrm{S}_{28}=14\left[36+\left(27 \times \frac{-5}{2}\right)\right]$

$=14\left[36-\frac{135}{2}\right]$

$\mathrm{S}_{28}=14\left(\frac{72-135}{2}\right)=14 \times\left(\frac{-63}{2}\right)=-441$

Question 20

(i) How many terms of the A.P. – 6,$-\frac{11}{2}$  – 5,… make the sum – 25?

(ii) Solve the equation 2 + 5 + 8 + … + x = 155.

Sol :

(i) Sum = -25

A.P. $=-6,-\frac{11}{2}-5, \ldots$

Here, $a=-6, d=\frac{-11}{2}+6=\frac{1}{2}$

Sum=-25

Let n term be added to get the sum -25

$\therefore \mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$-25=\frac{n}{2}\left[2 \times(-6)+(n-1)\left(\frac{1}{2}\right)\right]$

$-25 \times 2=n\left[-12+\frac{1}{2} n-\frac{1}{2}\right]$

$-50=n\left[\frac{-25}{2}+\frac{1}{2} n\right]$

$\frac{1}{2} n^{2}-\frac{25}{2} n+50=0$

$\Rightarrow n^{2}-25 n+100=0$

$\left\{\begin{array}{r}\because 100=-20 \times-5 \\ -25=-20-5\end{array}\right\}$

$\Rightarrow n^{2}-5 n-20 n+100=0$

$\Rightarrow n(n-5)-20(n-5)=0$

$\Rightarrow(n-5)(n-20)=0$

Either n-5=0, then n=5

or n-20=0, then n=20

$\therefore$ Number of terms are 5 or 20

(ii) Solve the equation 2+5+8+...+x=155

Here, a=2, d=5-2=3, l=x

Sum =155

l=a+(n-1)d

x=2+(n-1)3=2+3n-3

$\Rightarrow x=3 n-1$ ...(i)

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 155=\frac{n}{2}[2 \times 2+(n-1) \times 3]$

$\Rightarrow 155 \times 2=n[4+3 n-3]$

$\Rightarrow 310=n(3 n+1)=3 n^{2}+n$

$\therefore 3 n^{2}+n-310=0$

$3 n^{2}-30 n+31 n-310=0$

3n(n-10)+31(n-10)=0

(n-10)(3 n+31)=0

Either n-10=0, then n=10

or 3n+31=0, then 3n=-31 

$\Rightarrow n=\frac{-31}{3}$

which is not possible being negative

$\therefore n=10$

Now, x=3n-1$=3 \times 10-1=30-1=29$

[From (i)]

Question 21

If the third term of an A.P. is 5 and the ratio of its 6th term to the 10th term is 7 : 13, then find the sum of first 20 terms of this A.P.

Sol :

3rd term of an A.P. = 5

Ratio in 6th term and 10th term = 7 : 13

Find $S_{20}$

Let a be the first term and d be the common difference

$\therefore a_{3}=a+(n-1) d \Rightarrow a+(3-1) d=5$

$\Rightarrow a+2 d=5$..(i)

Similarly 

$a_{6}$=a+5d and $a_{10}$=a+9d

$\therefore \frac{a+5 d}{a+9 d}=\frac{7}{13}$

$ \Rightarrow 7 a+63 d=13 a+65 d$

$\Rightarrow 13 a+65 d-7 a-63 d=0$

$\Rightarrow 6 a+2 d=0 \Rightarrow 3 a+d=0$

$\Rightarrow d=-3 a$...(i)

From (i) and (ii),

a+2d=5

$\Rightarrow a+2(-3 a)=5 $

$\Rightarrow a-6 a=5$

$\Rightarrow-5 a=5 $

$\Rightarrow a=\frac{5}{-5}=-1$

and d=-3a$=-3 \times(-1)=3$

Now sum of first 20 terms

$=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{20}{2}[2 \times(-1)+(20-1) \times 3]$

=10[-2+57]

$=10 \times 55=550$

Question 22

In an A.P., the first term is 2 and the last term is 29. If the sum of the terms is 155, then find the common difference of the A.P.

Sol :

In an A.P.

First term (a) = 2

Last term (l) = 29

Sum of terms = 155

$l=a_{n}=a+(n-1) d$

$\Rightarrow 29=2+(n-1) d \Rightarrow 29-2=d(n-1)$

$\Rightarrow d(n-1)=27$..(i)

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$155=\frac{n}{2}[2 \times 2+27]=\frac{n}{2}[4+27]$

$155=\frac{31}{2} n $

$\Rightarrow n=\frac{155 \times 2}{31}=10$

d(n-1)=27

$\Rightarrow d(10-1)=27$

$d \times 9=27 \Rightarrow d=\frac{27}{9}=3$

Question 23

The sum of first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25th term.

Sol :

Sum of first 14 terms = 1505

First term (a) = 10

Find 25th term

$S_{14}=\frac{n}{2}[2 a+(n-1) d]$

$1505=\frac{14}{2}[2 \times 10+(14-1) d]$

$1505=7[20+13 d] \Rightarrow 20+13 d=\frac{1505}{7}$

$13 d=-20+215=195$

$d=\frac{195}{13}=15$

Now, $a_{25}=a+(n-1) d$

=10+(25-1)(15)=10+24(15)

=10+360=370

Question 24

Find the number of terms of the A.P. – 12, – 9, – 6, …, 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.

Sol :

A.P. -12, -9, -6,…, 21

If 1 is added to each term, find the sum of there terms

Here, a=-12, d=-9-(-12)=-9+12=3

Last term (l)=21

$\therefore l=a_{n}=a+(n-1) d$

$\Rightarrow 21=-12+(n-1) \times 3 $

$\Rightarrow 21+12=3(n-1)$

$\Rightarrow \frac{33}{3}=n-1 \Rightarrow 11=n-1$

n=11+1=12

Now, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{12}{2}[2 \times(-12)+(12-1) \times 3]$

$=6[-24+33]=6 \times 9=54$

By adding 1 to each term of the given A.P. the now sum will be $54+1 \times 12$

=54+12=66

Question 25

The sum of first n term of an A.P. is 3n² + 4n. Find the 25th term of this A.P.

Sol :

Sn = 3n² + 4n

Sn – 1 = 3(n – 1)² + 4(n – 1)

$=3\left(n^{2}-2 n+1\right)+4(n-1)$

$=3 n^{2}-6 n+3+4 n-4$

$=3 n^{2}-2 n-1$

Now, $a_{n}=\mathrm{S}_{n}-\mathrm{S}_{n-1}$

$=\left(3 n^{2}+4 n\right)-\left(3 n^{2}-2 n-1\right)$

$=3 n^{2}+4 n-3 n^{2}+2 n+1$

=6n+1

$a_{25}=6(25)+1=150+1=151$

Question 26

In an A.P., the sum of first 10 terms is – 150 and the sum of next 10 terms is – 550. Find the A.P.

Sol :

In an A.P.

Sum of first 10 terms = -150

Sum of next 10 terms = -550, A.P. = ?

Sum of first 10 terms = -150

$\mathrm{S}_{10}=\frac{n}{2}[2 a+(n-1) d]$

$-150=\frac{10}{2}[2 a+9 d]=5(2 a+9 d)$

$\Rightarrow 10 a+45 d=-150$..(i)

$S_{20}=S_{10}+S_{10}=-150-550=-700$

$=\frac{20}{2}[2 a+19 d]$

=10(2 a+19 d)

$\Rightarrow 20a+190d=-700$...(ii)

20a+90d=-300 [Multiplying (i) by 2]

Subtracting 100d=-400

$d=\frac{-400}{100}=-4$

From (i)

$10 a+45 \times(-4)=-150$

10a-180=-150

10a=-150+180=30

$a=\frac{30}{10}=3$

A.P. is 3,-1,-5,-9, ...

Question 27

The sum of first m terms of an A.P. is 4m² – m. If its nth term is 107, find the value of n. Also find the 21 st term of this A.P.

Sol :

Sm = 4m² – m

Sn = 4n² – n

and $S_{n-1}=4(n-1)^{2}-(n-1)$

$=4\left[n^{2}-2 n+1\right]-n+1$

$=4 n^{2}-8 n+4-n+1=4 n^{2}-9 n+5$

Now, $a_{n}=\mathrm{S}_{n}-\mathrm{S}_{n-1}$

$=4 n^{2}-n-4 n^{2}+9 n-5$

=8a-5

Now, $a_{n}=107$

$\therefore 8_{n}-5=107 $

$\Rightarrow 8 n=107+5=112$

$n=\frac{112}{8}=14$

and $a_{n}=8_{n}-5$

$a_{21}=8 \times 21-5=168-5=163$

Question 28

If the sum of first p, q and r terms of an A.P. are a, b and c respectively, prove that

$\frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)=0$

Sol :

Let the first term of A.P. be A and common difference be d.

Sum of the first p terms is

$\mathrm{S}_{p}=\frac{p}{2}[2 \mathrm{~A}+(p-1) d]=a$...(i)

Sum of the first q terms is

$S_{q}=\frac{q}{2}[2 A+(q-1) d]=b$..(ii)

Sum of the first r terms is

$S_{r}=\frac{r}{2}[2 A+(r-1) d]=c$..(iii)

Now, Multiply

(i) with $\frac{q-r}{p},$(ii) with $\frac{r-p}{q},$ (iii) with

$\frac{p-q}{r}$

$\therefore \frac{a}{p}(q-r)+\frac{b}{q}(r-p)+\frac{c}{r}(p-q)$

$=\frac{(q-r)}{2}[2 \mathrm{~A}+(p-1) d]+\frac{(r-p)}{2}[2 \mathrm{~A}+(q-1) d+\frac{(p-q)}{2}[2 \mathrm{~A}+(r-1) d]$

$=\frac{2 \mathrm{~A}}{2}(q-r+r-p+p-q)+\frac{d}{2}[(q-r)(p-1)+(r-p)(q-1)+(p-q)(r-1)]$

$=0+\frac{d}{2}[(q-r) p+(r-p) q+(p-q) r-q+r-r+p-p+q]$

$=\frac{d}{2}[p q-p r+r q-p q+p r-q r+0]$

$=\frac{d}{2} \times 0=0$

Which is the required result.

Question 29

A sum of Rs 700 is to be used to give 7 cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

What is the importance of an academic prise in students life? (Value Based)

Sol :

Total sum = Rs 700

Cash prizes to 7 students = 7 prize

Each prize is Rs 20 less than its preceding prize

d = -20, d = -20, n = 7

Now, $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$700=\frac{7}{2}[2 a+(7-1)(-20)]$

$\frac{700 \times 2}{7}=2 a-120 \Rightarrow 200=2 a-120$

$2 a=200+120=320 \Rightarrow a=\frac{320}{2}=160$

$\therefore$ Cash prizes will be ₹ 160, ₹ 140, ₹ 120, ₹ 100, ₹ 80, ₹ 60, ₹ 40

Question 30

Find the geometric progression whose 4th term is 54 and 7th term is 1458.

Sol :

In a G.P.

$a_{4}=54$

$a_{7}=1458$

Let a be the first term and r be the common difference

$\therefore a_{4}=a r^{n-1}=a r^{3}=54$

Similarly, $a r^{6}=1458$

Dividing, we get

$\frac{a r^{6}}{a r^{3}}=\frac{1458}{54} \Rightarrow r^{3}=27=(3)^{3}$

$\therefore r=3$

and $a r^{3}=54 \Rightarrow a \times 27=54$

$\Rightarrow a=\frac{54}{27}=2$

$\therefore a=2, r=3$

and G.P. is 2,6,18,54, ...

Question 31

The fourth term of a G.P. is the square of its second term and the first term is – 3. Find its 7th term.

Sol :

In G.P.

$a_{4}=\left(a_{2}\right)^{2}, a_{1}=-3$

$a_{4}=\left(a_{2}\right)^{2}, a_{1}=-3$

$\therefore a_{n}=a r^{n-1}$

$a_{4}=a r^{3}$

$a_{2}=a r$

$a r^{3}=(a r)^{2} \Rightarrow a r^{3}=a^{2} r^{2}$

$\Rightarrow r=a \Rightarrow a_{1}=-3$

$\therefore d=-3$

$\therefore a_{7}=a r^{7}-1=a r^{6}=-3 \times(-3)^{6}$

$=-3 \times 729=-2187$

Question 32

If the 4th, 10th and 16th terms of a G.P. are x, y and z respectively, prove that x, y and z are in G.P.

Sol :

In a G.P.

$a_{4}=x, a_{10}=y, a_{16}=z$

Show that x, y, z are in G.P.

Let a be the first term and r be the common ratio, then

$a_{n}=a r^{n-1}$

$a_{4}=a r^{4-1}=a r^{3}=x$

Similarly, $a_{10}=a r^{9}=y$

$a_{16}=a r^{15}=z$

x, y, z are in G.P.

if $y^{2}=x y$

$y^{2}=\left(a r^{9}\right)^{2}=a^{2} r^{18}$

$x z=a r^{3} \times a r^{15}=a^{2} r^{3+15}=a^{2} r^{18}$

$\therefore y^{2}=x y$

$\therefore x, y, z$ are in G.P.

Question 33

The original cost of a machine is Rs 10000. If the annual depreciation is 10%, after how many years will it be valued at Rs 6561 ?

Sol :

Original cost of machine = Rs 10000

Since, machine depreciates at the rate of 10%

on reducing the balance,

Value of machine after one year

$=10,000 \times \frac{90}{100}$

Value of the machine after two years

$=10,000\left(\frac{90}{100}\right) \times \frac{90}{100}$

$=10,000 \times\left(\frac{90}{100}\right)^{2}$

Thus, $10,000\left(\frac{90}{100}\right), \cdot 10,000\left(\frac{90}{100}\right)^{2} \ldots$ will

form a G.P. series with $a=10,000\left(\frac{90}{100}\right)$

and $r=\frac{9}{10}$

Let value of the machine after n years be ₹ 656

$\therefore \mathrm{T}_{n}=a r^{n-1}$

$\Rightarrow 6561=10,000\left(\frac{9}{10}\right) \times\left(\frac{9}{10}\right)^{n-1}$

$\Rightarrow \frac{6561}{10000}=\left(\frac{9}{10}\right)^{n}$

$\Rightarrow\left(\frac{9}{10}\right)^{4}=\left(\frac{9}{10}\right)^{n}$

$\Rightarrow n=4$

$\therefore n=4$

Hence, the effective life of the machine is 4 years.

Question 34

How many terms of the G.P. $3, \frac{3}{2}, \frac{3}{4}$ ,are needed to give the sum $\frac{3069}{512}$

Sol :

G.P.3, $\frac{3}{2}, \frac{3}{4}$

$S_{n}=\frac{3069}{512}$

Here, $a=3, r=\frac{1}{2}$

Let n be the number of terms

$\mathrm{S}_{n}=\frac{a\left(1-r^{n}\right)}{1-r}$

$\frac{3069}{512}=\frac{3\left[1-\left(\frac{1}{2}\right)^{n}\right]}{1-\frac{1}{2}}$

$=\frac{3\left[1-\left(\frac{1}{2}\right)^{n}\right]}{\frac{1}{2}}$

$\frac{3069}{512}=2 \times 3\left[1-\left(\frac{1}{2}\right)^{n}\right]$

$1-\left(\frac{1}{2}\right)^{n}=\frac{3069}{512 \times 6}=\frac{1023}{1024}$

$\left(\frac{1}{2}\right)^{n}=\frac{1024-1023}{1024}=\frac{1}{1024}$

$\begin{array}{l|l}2 & 1024 \\\hline 2 & 512 \\\hline 2 & 256 \\\hline 2 & 128 \\\hline 2 & 64 \\\hline 2 & 32 \\\hline 2 & 16 \\\hline 2 & 8 \\\hline 2 & 4 \\\hline 2 & 2 \\\hline & 1\end{array}$

$\left(\frac{1}{2}\right)^{n}=\left(\frac{1}{2}\right)^{10}$

Comparing, we get

n=10

Question 35

Find the sum of first n terms of the series : 3 + 33 + 333 + …

Sol :

Series is

3 + 33 + 333 + … n terms

= 3[1 + 11 + 111 +…n terms]

$=\frac{3}{9}[9+99+999+\ldots n$ terms $]$

$=\frac{3}{9}[(10-1)+(100-1)+(1000-1)+\ldots$n terms]

$=\frac{3}{9}[10+100+1000+\ldots n$ terms $-n \times 1]$

$=\frac{3}{9}\left[\left(\frac{a\left(r^{n}-1\right.}{r-1}\right)-n\right]$

$=\frac{3}{9}\left[\frac{10\left(10^{n}-1\right)}{10-1}-n\right]$

$=\frac{3}{9}\left[\frac{10}{9}\left(10^{n}-1\right)-9 n\right]$

$=\frac{3}{81}\left[10 \times 10^{n}-10-9 n\right]$

$=\frac{1}{27}\left[10^{n+1}-9 n-10\right]$

Question 36

Find the sum of the series 7 + 7.7 + 7.77 + 7.777 + … to 50 terms.

Sol :

The given sequence is 7, 7.7, 7.77, 7.777,…

Required sum $=S_{50}$

= 7 + 7.7 + 7.77 + … 50 terms

= 7(1 + 1.1 + 1. 11 + … 50 terms)

$=\frac{7}{9}[9+9.9+9.99+50$ terms ]

$=\frac{7}{9}(10-1)+1(10-0.1)+(10-0.01)+50$ terms)

$=\frac{7}{9}[(10+10+10+\ldots 50$ terms $)-(1+0.1+0.01+\ldots 50$ terms )]

$=\frac{7}{9}[500-$ Sum of G.P. of 50 terms with a=1, r=0.1]

$=\frac{7}{9}\left[500-\frac{1\left[1-(0.1)^{50}\right]}{1-0.1}\right]$

$=\frac{7}{9}\left[500-\frac{10}{9}\left(1-\frac{1}{10^{50}}\right)\right]$

$=\frac{7}{81}\left(4500-10+10^{-49}\right)$

$=\frac{7}{81}\left[4490+10^{-49}\right]$

Question 37

The inventor of chessboard was a very clever man. He asked the king, h reward of one grain of wheat for the first square, 2 grains for the second, 4 grains for the third, and so on, doubling the number of the grains for each subsequent square. How many grains would have to be given?

Sol :

In a chessboard, there are 8 x 8 = 64 squares

If a man put 1 grain in first square,

2 grains in second square,

4 grains in third square

and goes on upto the last square, i.e. 64th square

Therefore, 1 + 2 + 4 + 8 + 16 + … 64 terms

Here, a = 1, r = 2 and n = 64

$\mathrm{S}_{64}=\frac{a\left(r^{\prime \prime}-1\right)}{r-1}$

$=\frac{1\left(2^{64}-1\right)}{2-1}$

$=2^{64}-1$

ML Aggarwal Solution Class 10 Chapter 9 Arithmetic and Geometric Progressions MCQs

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 Q31 Q32 Q33

 MCQs

Question 1

The list of numbers – 10, – 6, – 2, 2, … is

(a) an A.P. with d = – 16

(b) an A.P with d = 4

(c) an A.P with d = – 4

(d) not an A.P

Sol :

-10, -6, -2, 2, … is

an A.P. with d = – 6 – (-10)

= -6 + 10 = 4

Ans (b)

Question 2

The 10th term of the A.P. 5, 8, 11, 14, … is

(a) 32

(b) 35

(c) 38

(d) 185

Sol :

10th term of A.P. 5, 8, 11, 14, …

{∵ a = 5, d = 3}

a + (n – 1)d = 5 + (10 – 1) x 3

= 5 + 9 x 3

= 5 + 27

= 32 

Ans (a)

Question 3

The 30th term of the A.P. 10, 7, 4, … is

(a) 87

(b) 77

(c) – 77

(d) – 87

Sol :

30th term of A.P. 10, 7, 4, … is

30th term = a + (n – 1)d

$\left\{\begin{aligned} \because a &=10 \\ d &=7-10=-3 \end{aligned}\right\}$

$=10+(30-1) \times(-3)$

$=10+29(-3)=10-87=-77$

Ans (c)

Question 4

The 11th term of the A.P. – 3, $-\frac{1}{2}, 2, \ldots$ is

(a) 28

(b) 22

(c) – 38

(d) – 48

Sol :

Given

$-3,-\frac{1}{2}, 2, \ldots$

$a=-3, d=-\frac{1}{2}-(-3)=-\frac{1}{2}+3=\frac{5}{2}$

11 th term $=a+(n-1) d$

$=-3+(11-1) \times \frac{5}{2}$

$=-3+10 \times \frac{5}{2}=-3+25=22$

Ans (b)

Question 5

The 4th term from the end of the A.P. – 11, – 8, – 5, …, 49 is

(a) 37

(b) 40

(c) 43

(d) 58

Sol :

4th term from the end of the A.P. -11, -8, -5, …, 49 is

Here, a = -11, d = -8 – (-11) = -8 + 11 = 3 and l = 49 .

$\therefore l=49=a+(n-1) d$

$\Rightarrow 49=-11+(n-1) \times 3$

$\Rightarrow 49-11=3(n-1)$

$\Rightarrow \frac{60}{3}=n-1 $

$\Rightarrow n=20+1=21$

Now, 4 th term from the end $=l-(n-1) d$ 

$=49-(4-1) \times 3=49-9=40$

Ans (b)

Question 6

The 15th term from the last of the A.P. 7, 10, 13, …,130 is

(a) 49

(b) 85

(c) 88

(d) 110

Sol :

15th term from the end of A.P. 7, 10, 13,…, 130

Here, a = 7, d = 10 – 7 = 3, l = 130

15th term from the end = l – (n – 1)d

= 130 – (15 – 1) x 3

= 130 – 42

= 88 

Ans (c)

Question 7

If the common difference of an A.P. is 5, then $a_{18}-a_{13}$ is

(a) 5

(b) 20

(c) 25

(d) 30

Sol :

Common difference of an A.P. (d) = 5

$a_{18}-a_{13}=a+17 d-a-12 d=5 d=5 \times 5=25$

Ans (c)

Question 8

In an A.P., if $a_{18}-a_{14}=32$ then the common difference is

(a) 8

(b) – 8

(c) – 4

(d) 4

Sol :

If $a_{18}-a_{14}=32$, then d = ?

(a + 17d) – a – 13d = 32

⇒ a + 17d – a – 13d = 32

⇒ 4d = 32

$\Rightarrow d \frac{32}{4}=8$

Ans (a)

Question 9

In an A.P., if d = – 4, n = 7, $a_n$ = 4, then a is

(a) 6

(d) 7

(c) 20

(d) 28

Sol :

In an A.P., d = -4, x = 7, $a_n$ = 4 then a = ?

$a_n$ = a(n – 1)d = 4

$a_7$ = a + (7 – 1)d = 4

⇒ a + 6d = 4

⇒ a + 6 x (-4) = 4

a – 24 = 4

⇒ a = 4 + 24 = 28

Ans (d)

Question 10

In an A.P., if a = 3.5, d = 0, n = 101, then a_n will be

(a) 0

(b) 3.5

(c) 103.5

(d) 104.5

Sol :

In an A.P.

a = 3.5, d= 0, n = 101, then an = ?

a_n = a₁₀₁ = a + (101 – 1)d

= 3.5 + 100d

= 3.5 + 100×0

= 3.5 + 0

= 3.5 (b)

Question 11

In an A.P., if a = – 7.2, d = 3.6, a_n = 7.2, then n is

(a) 1

(b) 3

(c) 4

(d) 5

Sol :

In an A.P.

a = – 7.2, d = 3.6, an = 7.2, n = ?

a_n = 7.2

a + (n – 1)d = 7.2

– 7.2 + (n – 1) 3.6 = 7.2

(n – 1) x 3.6 = 7.2 + 7.2 = 14.4

$(n-1)=\frac{14.4}{3.6}=4$

n=4+1=5

Ans (d)

Question 12

Which term of the A.P. 21, 42, 63, 84,… is 210?

(a) 9th

(b) 10th

(c) 11th

(d) 12th

Sol :

Which term of an A.P. 21, 42, 63, 84, … is 210

Let 210 be nth term, then

Here, a = 21, d = 42 – 21 =21

210 = a + (n – 1)d

210 = 21 + (n – 1) x 21

⇒ 210 – 21 = 21(n – 1)

$\Rightarrow \frac{189}{21}=n-1$

⇒ 9 = n – 1

⇒ n = 9 + 1 = 10

∴ It is 10th term.

Ans (b)

Question 13

If the last term of the A.P. 5, 3, 1, – 1,… is – 41, then the A.P. consists of

(a) 46 terms

(b) 25 terms

(c) 24 terms

(d) 23 terms

Sol :

Last term of an A.P. 5, 3, 1, -1, … is -41

Then A.P. will consist of ……. terms

Here, a = 5, d = 3 – 5 = – 2 and n =?

l = -41

l = -41

l = -41 = a + (n – 1 )d

-41 = 5 + (n – 1) (-2)

-41 – 5 = (n – 1) (-2)

$\Rightarrow \frac{-46}{-2}=n-1$

⇒ n – 1 = 23

⇒ n = 23 + 1 = 24

A.P. consists of 24 terms. 

Ans (c)

Question 14

If k – 1, k + 1 and 2k + 3 are in A.P., then the value of k is

(a) – 2

(b) 0

(c) 2

(d) 4

Sol :

k – 1, k + 1 and 2k + 3 are in A.P.

2(k+ 1) = (k – 1) + (2k + 3)

⇒ 2k + 2 = k – 1 + 2k + 3

⇒ 2k + 2 – 3k + 2

⇒ 3k – 2k = 2 – 2

⇒ k = 0 (b)

Question 15

The 21st term of an A.P. whose first two terms are – 3 and 4 is

(a) 17

(b) 137

(c) 143

(d) – 143

Sol :

First two terms of an A.P. are – 3 and 4

a = -3, d = 4 – (-3) = 4 + 3 = 7

21st term = a + 20d

= -3 + 20(7)

= -3 + 140

= 137

Ans (b)

Question 16

If the 2nd term of an A.P. is 13 and the 5th term is 25, then its 7th term is

(a) 30

(b) 33

(c) 37

(d) 38

Sol :

In an A.P.

2nd term = 13 ⇒ a + d = 13 …(i)

5th term = 25 ⇒ a + 4d = 25 …(ii)

Subtracting (i) and (ii),

3d = 12

$\Rightarrow d=\frac{1}{3}$

Substitute the value of d in eq. (i), we get

a = 13 – 4 = 9

7th term = a + 6d = 9 + 6 x 4 = 9 + 24 = 33 

Ans (b)

Question 17

If the first term of an A.P. is – 5 and the common difference is 2, then the sum of its first 6 terms is

(a) 0

(b) 5

(c) 6

(d) 15

Sol :

First term (a) of an A.P. = -5

Common difference (d) is 2

Sum of first 6 terms $=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{6}{2}[2 \times(-5)+(6-1) \times 2]$

$=3[-10+5 \times 2]$

$=3 \times[-10-10]$

$=3 \times 0=0$

Ans (a)

Question 18

The sum of 25 terms of the A.P. $-\frac{2}{3},-\frac{2}{3},-\frac{2}{3}$ is 

(a) 0

(b) $-\frac{2}{3}$

(c) $-\frac{50}{3}$

(d) -50

Sol :

Sum of 25 terms of an A.P. $-\frac{2}{3},-\frac{2}{3},-\frac{2}{3}$ is

$=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{25}{2}\left[2 \times \frac{(-2)}{3}+(25-1) \times 0\right]$

$=\frac{25}{2}\left[\frac{-4}{3}\right]=\frac{-50}{3}$

Ans (c)

Question 19

In an A.P., if $a=1, a_{n}=20$ and $S_{n}=399,$ then n is

(a) 19

(b) 21

(c) 38

(d) 42

Sol :

In an A.P. $\mathrm{a}=1, \mathrm{a}_{n}=20, \mathrm{~S}_{n}=399, \mathrm{n}$ is ?

$a_{n}=a+(n-1) d=20$

$1+(n-1) d=20$

$(n-1) d=20-1=19 \ldots(i)$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$399=\frac{n}{2}[2 \times 1+19]$ [From (i)]

$\Rightarrow 399=\frac{n}{2} \times 21$

$n=\frac{399 \times 2}{21}=38$

$\therefore n=38$

Ans (c)

Question 20

In an A.P., if $a=-5,l=21$. and $S_{n}=200$, then n is equal to

(a) 50

(b) 40

(c) 32

(d) 25

Sol :

In an A.P.

$a=-5,l=21, S_{n}=200, n=?$

l = a + (n – 1)d = -5 + (n – 1 )d

21 = -5 + (n – 1)d

$\Rightarrow(n-1) d=21+5=26$..(i)

$S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$200=\frac{n}{2}[2 \times(-5)+26]$ [From (i)]

$400=n[-10+26]=n(16)$

$n=\frac{400}{16}=25$

$\therefore n=25$

Ans (d)

Question 21

In an A.P., if a=3 and $S_{8}=192$, then d is

(a) 8

(b) 7

(c) 6

(d) 4

Sol :

In an A.P.

$a=3, S_{8}=192, d=?$

$\mathrm{S}_{8}=\frac{n}{2}[2 a+(n-1) d]$

$192=\frac{8}{2}[2 \times 3+(8-1) d]$

$\Rightarrow 192=4[6+7 d]$

$\frac{192}{4}=6+7 d $

$\Rightarrow 48=6+7 d$

7d=48-6

7d=42

$d=\frac{42}{7}=6$

Ans (c)

Question 22

The sum of first five multiples of 3 is

(a) 45

(b) 55

(c) 65

(d) 75

Sol :

First 5 multiples of 3 :

3, 6, 9, 12, 15

Here, a = 3, d = 6 – 3 = 3

$\therefore \mathrm{S}_{5}=\frac{n}{2}[2 a+(n-1) d]$

$\therefore \mathrm{S}_{5}=\frac{n}{2}[2 a+(n-1) d]$

Ans (a)

Question 23

The number of two digit numbers which are divisible by 3 is

(a) 33

(b) 31

(c) 30

(d) 29

Sol :

Two digit number which are divisible by 3 is 12, 15, 18, 21, … 99

Here, a = 12, d = 3, l = 99

$1=a_{n}=a+(n-1) d$

⇒ 12 + (n – 1) x 3 = 99

⇒ (n – 1)3 = 99 – 12 = 87

$\Rightarrow n-1=\frac{87}{3}=29$

⇒ n = 29 + 1 = 30 

Ans (c)

Question 24

The number of multiples of 4 that lie between 10 and 250 is

(a) 62

(b) 60

(c) 59

(d) 55

Sol :

Multiples of 4 lying between 10 and 250 12, 16, 20, 24, …, 248

Here, a = 12, d = 16 – 12 = 4, l = 248

$l=a_{n}=a+(n-1) d$

$248=12+(n-1) \times 4$

$\Rightarrow 248-12=4(n-1)$

$\Rightarrow \frac{236}{4}=n-1 $

$\Rightarrow n-1=59$

$\therefore n=59+1=60$

Ans (b)

Question 25

The sum of first 10 even whole numbers is

(a) 110

(b) 90

(c) 55

(d) 45

Sol :

Sum of first 10 even whole numbers

Even numbers are 0, 2, 4, 6, 8, 10, 12, 14, 16, 18

Here, a = 0, d = 2, n = 10

$\mathrm{S}_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 0+(10-1) \times 2]$

$=5[0+18]=90$

Ans (b)

Question 26

The list of number $\frac{1}{9}, \frac{1}{3}, 1,-3, \ldots$ is a

(a) GP. with r=-3

(b) G.P. with $r=-\frac{1}{3}$

(c) GP. with r = 3

(d) not a G.P.

Sol:

The given list of numbers

$\frac{1}{9}, \frac{1}{3}, 1,-3, \ldots$

Here, $a=\frac{1}{9}, r=-\frac{1}{3} \div \frac{1}{9}=-\frac{1}{3} \times \frac{9}{1}=-3$

It is a G.P. with r=-3

Ans (a)

Question 27

The 11 th of the G.P. $\frac{1}{8},-\frac{1}{4}, 2,-1, \ldots$ is

(a) 64

(b) – 64

(c) 128

(d) – 128

Sol :

11th of the G.P.

$\frac{1}{8},-\frac{1}{4}, 2,-1, \ldots$ is

Here $a=\frac{1}{8}, r=-\frac{1}{4} \div \frac{1}{8}=-\frac{1}{4} \times \frac{8}{1}=-2$

$\therefore a_{11}=a r^{n-1}=\frac{1}{8}(-2)^{10}$

$=\frac{1}{2^{3}} \times(-1)^{10} \times 2^{10}=1 \times 2^{10-3}=2^{7}$

=128

Ans (c)

Question 28

The 5th term from the end of the G.P. 2, 6, 18, …, 13122 is

(a) 162

(b) 486

(c) 54

(d) 1458

Sol :

5th term from the end of the G.P. 2, 6, 18, …, 13122 is

Here, $a=2, r=\frac{6}{2}=3,1=13122$

$\therefore$ 5th term from the end $=l\left(\frac{1}{r}\right)^{m-n}$

Here, $l=a_{n}=a r^{n-1}$

$13122=2(3)^{n-1}$

$\Rightarrow \frac{13122}{2}=3^{(n-1)}$

$\Rightarrow 6561=3^{(n-1)}$

$\begin{array}{l|l}3 & 6561 \\\hline 3 & 2187 \\\hline 3 & 729 \\\hline 3 & 243 \\\hline 3 & 81 \\\hline 3 & 27 \\\hline 3 & 9 \\\hline 3 & 3 \\\hline & 1\end{array}$

$\Rightarrow(3)^{8}=3^{n-1}$

Comparing, we get

$n-1=8 \Rightarrow n=9$

5th term from the end 

$=l\left(\frac{1}{r}\right)^{m-n}$

$=13122\left(\frac{1}{3}\right)^{9-5}$

$=13122 \times\left(\frac{1}{3}\right)^{4}=\frac{13122}{3 \times 3 \times 3 \times 3}$

$ \begin{array}{l}81\overline{)13122(}162\\\phantom{81)}81\\\phantom{81)}\overline{502}\\\phantom{81)}486\\\phantom{81)}\overline{\phantom{4}162}\end{array}$

=162

Ans (a)

Question 29

If k, 2(k + 1), 3(k + 1) are three consecutive terms of a G.P., then the value of k is

(a) – 1

(b) – 4

(c) 1

(d) 4

Sol :

k, 2(k + 1), 3(k + 1) are in G.P.

[2(k + 1)]² = k x 3(k + 1)

⇒ 4(k + 1)² = 3k(k + 1)

⇒ 4 (k + 1) = 3 k

(Dividing by k + 1 if k + 1 ≠ 0) 4

⇒ 4k + 4 = 3k

⇒ 4k – 3k = -4

⇒ k = -4 

Ans (b)

Question 30

Which term of the G.P. 18, – 12, 8, … is $\frac{512}{729}$ ?

(a) 12th

(b) 11th

(c) 10th

(d) 9th

Sol :

Which term of the G.P.

$18,-12,8, \ldots \frac{512}{729}$

Let it be nth term

Here, $a=18, d=\frac{-12}{18}=\frac{-2}{3}$

$\therefore a_{n}=a r^{n-1}$

$\frac{512}{729}=18\left(\frac{-2}{3}\right)^{n-1}$

$ \Rightarrow \frac{512}{729} \times \frac{1}{18}=\left(\frac{-2}{3}\right)^{n-1}$

$\frac{256}{729 \times 9}=\left(\frac{-2}{3}\right)^{n-1}$

$ \Rightarrow\left(\frac{-2}{3}\right)^{8}=\left(\frac{-2}{3}\right)^{n-1}$

Comparing, n-1=8

$\Rightarrow n=8+1=9$

$\therefore$ It is 9 th term.

Ans (d)

Question 31

The sum of the first 8 terms of the series 1 + √3 + 3 + … is

Sol :

Sum of first 8 terms of 1 + √3 + 3 + … is

Here $a=1, r=\frac{\sqrt{3}}{1}=\sqrt{3}, n=8$

$S_{n}=\frac{a\left(r^{n}-1\right)}{r-1}$

$=\frac{1\left[(\sqrt{3})^{8}-1\right]}{\sqrt{3}-1}$

$=\frac{81-1}{\sqrt{3}-1}=\frac{80}{\sqrt{3}-1}$

$=\frac{80(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)}$

(Rationalising the denominator)

$\frac{80(\sqrt{3}+1)}{3-1}=\frac{80(\sqrt{3}+1)}{2}$

$=40(\sqrt{3}+1)$

Ans (a)

Question 32

The sum of first 6 terms of the G.P. 1, $-\frac{2}{3}, \frac{4}{9}, \ldots$ is

(a) $-\frac{133}{243}$ 3

(b) $\frac{133}{243}$

(c) $\frac{793}{1215}$

(d) none of these

Sol :

Sum of first 6 terms of G.P.

$1,-\frac{2}{3}, \frac{4}{9}, \ldots$

Here, $a=1, r=-\frac{2}{3}, n=6$

$\mathrm{S}_{6}=\frac{a\left(1-r^{n}\right)}{1-r}$

$=\frac{1\left[1-\left(\frac{2}{3}\right)^{6}\right]}{1+\frac{2}{3}}$

$=\left[1-\frac{64}{729}\right] \frac{3}{5}$

$=\frac{729-64}{729} \times \frac{3}{5}$

$=\frac{665}{243 \times 5}=\frac{133}{243}$

Ans (b)

Question 33

If the sum of the GP., 1,4, 16, … is 341, then the number of terms in the GP. is

(a) 10

(b) 8

(c) 6

(d) 5

Sol :

The sum of G.P. 1, 4, 16, … is 341

Let n be the number of terms,

Here, $a=1, r=\frac{4}{1}=4, \mathrm{~S}_{n}=341$

$341=\frac{a\left(r^{n}-1\right)}{r-1}=\frac{1\left(4^{n}-1\right)}{4-1}=\frac{4^{n}-1}{3}$

$341 \times 3=4^{n}-1 \Rightarrow 1023=4^{n}-1$

$\Rightarrow 1023+1=4^{n} \Rightarrow 4^{n}=1024$

$4^{n}=4^{5}$

$\therefore n=5$

Ans (d)

ML Aggarwal Solution Class 10 Chapter 9 Arithmetic and Geometric Progressions Exercise 9.4

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25 Q26 Q27 Q28 Q29 Q30 Q31

 Exercise 9.4

Question 1

Can 0 be a term of a geometric progression?

Sol :

No, 0 is not a term of geometric progression

Question 2

(i) Find the next term of the list of numbers $\frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \ldots$

(ii) Find the next term of the list of numbers  $\frac{3}{16},-\frac{3}{8}, \frac{3}{4},-\frac{3}{2}, \ldots$

(iii) Find the 15th term of the series $\sqrt{3}+\frac{1}{\sqrt{3}}+\frac{1}{3 \sqrt{3}}+\cdots$

(iv) Find the nth term of the list of numbers $\frac{1}{\sqrt{2}},-2,4 \sqrt{2},-16, \ldots$

(v) Find the 10th and nth terms of the list of numbers 5, 25, 125, …

(vi) Find the 6th and the nth terms of the list of numbers $\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \ldots$

(vii) Find the 6th term from the end of the list of numbers 3, – 6, 12, – 24, …, 12288.

Sol :

Given :

$\Rightarrow \frac{1}{6}, \frac{1}{3}, \frac{2}{3}, \ldots$

Here, $a=\frac{1}{6}, r=\frac{1}{3} \div \frac{1}{6}=\frac{1}{3} \times \frac{6}{1}=2$

$\therefore$ Next term $=\frac{2}{3} \times 2=\frac{4}{3}$

(ii) $\frac{3}{16},-\frac{3}{8}, \frac{3}{4},-\frac{3}{2}, \ldots$

Here, $a=\frac{3}{16}, r=\frac{-3}{8} \div \frac{3}{16}=\frac{-3}{8} \times \frac{16}{3}=-2$

$\therefore$ Next term $=\frac{-3}{2} \times(-2)=3$

(iii) $\sqrt{3}+\frac{1}{\sqrt{3}}+\frac{1}{3 \sqrt{3}}+\ldots$

Here, $a=\sqrt{3}, r=\frac{1}{\sqrt{3}} \div \sqrt{3}=\frac{1}{\sqrt{3}} \times \frac{1}{\sqrt{3}}$

$=\frac{1}{3}$

$\therefore a_{15}=a r^{n-1}=\sqrt{3}\left(\frac{1}{3}\right)^{15-1}$

$=\sqrt{3} \times\left(\frac{1}{3}\right)^{14}=\sqrt{3} \times \frac{1}{3^{14}}$

(iv) $\frac{1}{\sqrt{2}},-2,4 \sqrt{2},-16, \ldots$

Here, $a=\frac{1}{\sqrt{2}}, r=-2 \div \frac{1}{\sqrt{2}}=-2 \times \sqrt{2}=-2 \sqrt{2}$

$a_{n}=a r^{n-1}=\frac{1}{\sqrt{2}} \times(-2 \sqrt{2})^{n-1}$

$=\frac{1}{\sqrt{2}} \times(-1)^{n-1} \times\left[(\sqrt{2})^{2} \times \sqrt{2}\right]^{n-1}$

$=(-1)^{n-1} \times \frac{1}{\sqrt{2}} \times\left[(\sqrt{2})^{3}\right]^{n-1}$

$=(-1)^{n-1} \times \frac{1}{\sqrt{2}} \times(\sqrt{2})^{3 n-3}$

$=(-1)^{n-1}(\sqrt{2})^{3 n-3-1}$

$=(-1)^{n-1}(\sqrt{2})^{3 n-4}$

$=(-1)^{n-1} \times 2^{\frac{3 n-4}{2}}$

(v) 5,25,125,...

Here, a=5, $r=25 \div 5=5$

$a_{10}=a r^{n-1}=5 \times(5)^{10-1}$

$=5 \times 5^{9}=5^{9+1}=5^{10}$

$a_{n}=a r^{n-1}=5 \times 5^{n-1}=5^{n-1+1}=5^{n}$

(vi) $\frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \ldots$

Here, $a=\frac{3}{2}, r=\frac{3}{4} \div \frac{3}{2}=\frac{3}{4} \times \frac{2}{3}=\frac{1}{2}$

$\therefore a_{n}=a r^{n-1}=\frac{3}{2} \times \frac{1}{2}^{n-1}$

$=3 \times \frac{1}{2} \times\left(\frac{1}{2}\right)^{n-1}=3 \times\left(\frac{1}{2}\right)^{n-1+1}$

$=3 \times\left(\frac{1}{2}\right)^{n}=\frac{3}{2^{n}}$

$a_{6}=\frac{3}{2^{n}}=\frac{3}{2^{6}}=\frac{3}{64}$

(vii) 3,-6,12,-24,..., 12288

6th term from the end of the list

Here, $a=3, r=-6 \div 3=-2, l=12288$

Now. 6th term from the end

$=l \times\left(\frac{1}{r}\right)^{n-1}=12288 \times\left(\frac{1}{-2}\right)^{6-1}$

$=12288 \times \frac{1}{(-2)^{5}}=\frac{12288}{-32}=-384$

Question 3

Which term of the G.P.

(i) 2, 2√2, 4, … is 128?

(ii) $1, \frac{1}{3}, \frac{1}{9}, \ldots, 28 \text{is} \frac{1}{243} ?$

(iii) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683} ?$

Sol :

Given

(i) 2, 2√2, 4, … is 128?

Here $a=2, r=\frac{2 \sqrt{2}}{2}=\sqrt{2}, l=128$

Let 128 be the n th term, then

$a_{n}=128=a r^{n-1}$

$\Rightarrow 128=2(\sqrt{2})^{n-1} \Rightarrow 2(\sqrt{2})^{n-1}=2^{7}$

$(\sqrt{2})^{n-1}=2^{7-1}=2^{6}$

$(\sqrt{2})^{n-1}=(\sqrt{2})^{12}$

Comparing, we get n-1=12 

$\Rightarrow n=12+1=13$

$\therefore 128$ is the 13 th term

(ii) $1, \frac{1}{3}, \frac{1}{9}, \ldots$ is $\frac{1}{243}$

Here, $a=1, r=\frac{1}{3} \div 1=\frac{1}{3}, l=\frac{1}{243}$

Let $\frac{1}{243}$ is the n th term, then

$a_{n}=\frac{1}{243}=a r^{n-1}=1 \times\left(\frac{1}{3}\right)^{n-1}$

$\Rightarrow\left(\frac{1}{3}\right)^{n-1}=\left(\frac{1}{3}\right)^{5}$

Comparing, we get

n-1=5 

$\Rightarrow n=5+1=6$

$\therefore \frac{1}{243}$ is the 6 th term

(iii) $\frac{1}{3}, \frac{1}{9}, \frac{1}{27}, \ldots$ is $\frac{1}{19683}$

Here, $a=\frac{1}{3}, r=\frac{1}{9} \div \frac{1}{3}$

$=\frac{1}{9} \times \frac{3}{1}=\frac{1}{3}, l=\frac{1}{19683}$

Let $\frac{1}{19683}$ is the $n$ th term, then

$a_{n}=\frac{1}{19683}=a r^{h-1}=\frac{1}{3}\left(\frac{1}{3}\right)^{n-1}$

$=\frac{1}{3}^{n-1+1}=\left(\frac{1}{3}\right)^{n}$

$\Rightarrow \left(\frac{1}{3}\right)^{n}=\left(\frac{1}{3}\right)^{9}$

Comparing, we get

$\therefore n=9$

$\begin{array}{l|l} 3 & 19683 \\ \hline 3 & 6561 \\ \hline 3 & 2187 \\ \hline 3 & 729 \\ \hline 3 & 243 \\ \hline 3 & 81 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$

Hence, $\frac{1}{19683}$ is the 9 th term

Question 4

Which term of the G.P. 3, – 3√3, 9, – 9√3, … is 729 ?

Sol :

G.P. 3, -3√3, 9, – 9√3, … is 729 ?

Here $a=3, r=\frac{-3 \sqrt{3}}{3}=\sqrt{-3}, l=729$

Let 729 is the $n$ th term, then

$a_{n}=729=a r^{n-1}=3(-\sqrt{3})^{n-1}$

$\Rightarrow \frac{729}{3}=(-\sqrt{3})^{n-1} \Rightarrow 243=(-\sqrt{3})^{n-1}$

$\Rightarrow(-\sqrt{3})^{10}=(-\sqrt{3})^{n-1}$

Comparing, we get

$n-1=10 \Rightarrow n=10+1=11$

$\therefore 729$ is the 11 th term

Question 5

Determine the 12th term of a G.P. whose 8th term is 192 and common ratio is 2.

Sol :

In a G.P.

$a_8$ = 192 and r = 2

Let a be the first term and r be the common ratio then.

$a_{8}=a r^{n-1} \Rightarrow 192=a(2)^{n-1}=a 2^{8-1}=a 2^{7}$

$\Rightarrow a=\frac{192}{2^{7}}=\frac{192}{128}=\frac{3}{2}$

$\therefore a_{12}=a(r)^{n-1}$

$\Rightarrow a_{12}=\frac{3}{2}(2)^{12-1}=\frac{3}{2} \times 2^{11}$

$=\frac{3}{2} \times 2048=3072$

$\therefore \mathrm{a}_{12}=3072$

Question 6

In a GP., the third term is 24 and 6th term is 192. Find the 10th term

Sol :

In a GP.

$a_{3}=24$ and $a_{6}=192, a_{10}=?$

Let a be the first term and r be the common ratio, therefore

$a_{6}=a r^{6-1}=a r^{5}=192 \quad\left\{\because a_{n}=a r^{n-1}\right\}$

$a_{3}=a r^{3-1}=a r^{2}=24$

Dividing, we get

$\frac{a r^{5}}{a r^{2}}=\frac{192}{24} \Rightarrow r^{3}=8=(2)^{3}$

$\therefore r=2$

$\quad$ Now, $a r^{2}=24 \Rightarrow a \times 2^{2}=24$

$\Rightarrow a=\frac{24}{2^{2}}=\frac{24}{4}=6$

$\therefore \quad a=6$

Now, $a_{10}=a r^{10-1}=a r^{9}$

$=6 \times(2)^{9}=6 \times 512=3072$

Question 7

Find the number of terms of a G.P. whose first term is $\frac{3}{4}$, common ratio is 2 and the last term is 384.

Sol :

First term of a G.P. (a) $=\frac{3}{4}$

and common ratio (r) = 2

Last term = 384

Let number of terms is n

then $a_{n}=a r^{n-1}$

$\Rightarrow 384=\frac{3}{4}(2)^{n-1}$

$\Rightarrow 2^{n-1}=\frac{384 \times 4}{3}=512=2^{9}$

$\therefore n-1=9 $

$\Rightarrow n=9+1=10$

$\therefore$ Number of terms in G.P. =10

Question 8

Find the value of x such that

(i) $-\frac{2}{7}, x ,-\frac{7}{2}$  are three consecutive terms of a G.P.

(ii) x + 9, x – 6 and 4 are three consecutive terms of a G.P.

(iii) x, x + 3, x + 9 are first three terms of a G.P. Sol. Find the value of x

Sol :

Find the value of x

(i) $-\frac{2}{7}, x_{3}-\frac{7}{2}$  are three consecutive terms of a G.P.

$\therefore x^{2}=\frac{-2}{7} \times \frac{-7}{2}=1=(\pm 1)^{2}$

$\therefore x=+1$

(ii) x+9, x-6 and 4 are three consecutive terms of a G.P., then

$(x-6)^{2}=(x+9) \times 4$

$\Rightarrow x^{2}-12 x+36=4 x+36$

$\Rightarrow x^{2}-12 x-4 x+36-36=0$

$\Rightarrow x^{2}-16 x=0 \Rightarrow x(x-16)=0$

Either x-16=0, then x=16

or x=0

$\therefore x=0,16$

(iii) x, x+3, x+9 are first three terms of a G.P.

$\therefore(x+3)^{2}=x(x+9)$

$\Rightarrow x^{2}+6 x+9=x^{2}+9 x$

9=9x-6x=3x

$\therefore x=\frac{9}{3}=3$

Question 9

If the fourth, seventh and tenth terms of a G.P. are x, y, z respectively, prove that x, y, z are in G.P.

Sol :

In a G.P.

$a_{4}=x, a_{7}=y, a_{10}=z$

Let a be the first term and r be the common

ratio, therefore 

$a_{4}=a r^{n-1}=a r^{4-1}=a r^{3}=x$

Similarly, 

$a_{7}=a r^{6}=y$

$a_{10}=a r^{9}=z$

If x, y and z are in G.P., then

$y^{2}=x z$

Now, $x z=a r^{3} \times a r^{9}=a^{2} r^{3+9}=a^{2} r^{12}$

$y^{2}=\left(a r^{6}\right)^{2}=a^{2} \cdot r^{12}$

$\because$ L.H.S. $=$ R.H.S.

$\therefore x, y$ and $z$ are in G.P.

Question 10

The 5th, 8th and 11th terms of a G.P. are p, q and s respectively. Show that q² = ps.

Sol :

In a G.P.

$a_{5}=p, a_{8}=q$ and $a_{11}=s$

To show that $q^{2}=p x$

Let a be the first term and r be the common

$a_{5}=a r^{n-1}=a r^{5-1}=a r^{4}=p$

Similarly, $a_{8}=a r^{7}=q$ and

$a_{11}=a r^{10}=s$

$q^{2}=\left(a r^{7}\right)^{2}=a r^{14}$

and $p x=a r^{4} \times a r^{10}=a^{2} r^{4+10}=a^{2} r^{14}$

Hence, $q^{2}=p s$

Question 11

If a, b, c are in G.P., then show that a², b², c² are also in G.P.

Sol :

a, b, c are in G.P.

Show that a², b², c² are also in G.P

∵ a, b, c are in G.P., then

b² = ac …(i)

a², b², c² will be in G.P.

if (b²)² = a² x c²

⇒ (ac)² = a²c² [From (i)]

⇒ a²c² = a²c² which is true.

Hence proved.

Question 12

If a, b, c are in A.P., then show that $3^{a}, 3^{b}, 3^{c}$ are in G.P.

Sol :

a, b and c are in A.P.

Then, 2b = a + c

Now, $3^{a}, 3^{b}, 3^{c}$ will be in G.P.

if $\left(3^{b}\right)^{2}=3^{a} \cdot 3^{c}$

if $3^{2 b}=3^{a+c}$

Comparing, we get

if 2b = a + c

Which are in A.P. is given

Question 13

If a, b, c are in A.P., then show that $10^{a x+10}, 10^{b x+10}, 10^{c x+10}, x \neq 0$, are in G.P.

Sol :

a, b, c are in A.P.

To show that are in G.P. $10^{a x+10}, 10^{b x+10}, 10^{c x+10}, x \neq 0$

∵ a, b, c are in A.P.

$\therefore 2 b=a+c$...(i)

Now 

$\left(10^{\alpha x+10}\right),\left(10^{b x+10}\right),\left(10^{c x+10}\right)$ will be in

G.P. if $\left(10^{b x+10}\right)^{2}=\left(10^{a x+10}\right) \times\left(10^{c x+10}\right)$

if $10^{2 b x+20}=10^{a x+10+c x+10}$

if $10^{2 b x+20}=10^{a x+c x+20}$

Comparing, if 2bx+20=ax+cx+20

if 2bx=ax+cx

if 2b=a+c

Which is given

Question 14

If $a, a^{2}+2$ and $a^{3}+10$ are in G.P., then find the values(s) of a.

Sol :

$a, a^{2}+2$ and $a^{3}+10$ are in G.P.

$\because\left(a^{2}+2\right)^{2}=a\left(a^{3}+10\right)$

$\Rightarrow a^{4}+4 a^{2}+4=a^{4}+10 a$

$\Rightarrow 4 a^{2}-10 a+4=0$

$\Rightarrow 2 a^{2}-5 a+2=0$

$\Rightarrow 2 a^{2}-a-4 a+2=0$

⇒a(2 a-1)-2(2 a-1)=0

⇒(2 a-1)(a-2)=0

Either 2a-1=0, then 2a=1 $\Rightarrow a=\frac{1}{2}$

or a-2=0, then a=2

Hence a=2 or $\frac{1}{2}$

Question 15

If k, 2k + 2, 3k + 3, … are in G.P., then find the common ratio of the G.P.

Sol :

k, 2k + 2, 3k + 3, … are in G.P.

then $(2 k+2)^{2}=k(3 k+3)$

$\Rightarrow 4 k^{2}+8 k+4=3 k^{2}+3 k$

$\Rightarrow 4 k^{2}+8 k+4-3 k^{2}-3 k=0$

$\Rightarrow k^{2}+5 k+4=0$

$\Rightarrow k^{2}+k+4 k+4=0$

⇒k(k+1)+4(k+1)=0

⇒(k+1)(k+4)=0

Hence , k=-1 or -4

Question 16

The first and the second terms of a GP. are $x^{4}$ and $x^{m}$ . If its 8th term is $x^{52}$, then find the value of m.

Sol :

In a G.P.,

First term $\left(a_{1}\right)=x^{-4}$ …(i)

Second term $\left(a_{2}\right)=x^{m}$

Eighth term $\left(a_{8}\right)=x^{52}$

$r=\frac{a_{2}}{a_{1}}=\frac{x^{m}}{x^{-4}}=x^{m+4}$...(ii)

Now, $a_{8}=a r^{n-1}=a r^{8-1}=a r^{7}$

$x^{52}=x^{-4} \times r^{7}=x^{-4} \times x^{7(m+4)}$

$x^{52}=x^{-4+7 m+28}=x^{7 m+24}$

Comparing 

$52=7 m+24 \Rightarrow 7 m=52-24=28$

$\Rightarrow m=\frac{28}{7}=4$

Hence m=4

Question 17

Find the geometric progression whose 4th term is 54 and the 7th Term is 1458.

Sol :

In a G.P.,

4 th term $\left(a_{4}\right)=54$

and 7 th term $\left(a_{7}\right)=1458$ Let $a$ be the first term and r be the common ratio, then $a r^{3}=54$ and $a r^{6}=1458$

Dividing

$\frac{a r^{6}}{a r^{3}}=\frac{1458}{54}$

$ \Rightarrow r^{3}=27=\left(3^{3}\right)$

$\therefore r=3$

Now, $a r^{3}=54$

$\Rightarrow a \times 27=54 \Rightarrow a=\frac{54}{27}=2$

Hence G.P. is

2,6,18,54, ....

Question 18

The fourth term of a GP. is the square of its second term and the first term is – 3. Determine its seventh term.

Sol :

In a GP.

$a_{n}$ is square of $a_{2}$ i.e. 

$a n=\left(a_{2}\right)^{2}$ $a_{1}=-3$

Let $a$ be the first term and $r$ be the common ratio, then $\quad a_{4}=a r^{n-1}=a r^{4-1}=a r^{3}$

and $a_{2}=a r^{2}-1=a r$

$\therefore a r^{3}=(a r)^{2}$

$\Rightarrow a r^{3}=a^{2} r^{2}$

$\frac{r^{3}}{r^{2}}=\frac{a^{2}}{a}=r=a=-3 \quad\left(\because a_{1}=-3\right)$

Now, $a_{7}=a r^{7-1}=a r^{6}=(-3)(-3)^{6}=(-3)^{7}$

=-2187

Question 19

The sum of first three terms of a G.P. is $\frac{39}{10}$  and their product is 1. Find the common ratio and the terms.

Sol :

Sum of first three terms of G.P. $=\frac{39}{10}$ and their product =1

Let a be the first term and r be the common ratio, then

Let $\frac{a}{r}, a, a r$ be the three terms of $\mathrm{G}$. ., then

$\frac{a}{r}+a+a r=\frac{39}{10}$ and $\frac{a}{r} \times a \times a r=1$

$\Rightarrow a\left(\frac{1}{r}+1+r\right)=\frac{39}{10}$ and $a^{3}=1 \Rightarrow a=1$

Substitute the value of $a$

$\therefore 1\left(\frac{1}{r}+1+r\right)=\frac{39}{10}$

$\frac{1+r+r^{2}}{r}=\frac{39}{10}$

$10+10 r+10 r^{2}=39 r$

$\Rightarrow 10 r^{2}+10 r-39 r+10=0$

$\Rightarrow 10 r^{2}-29 r+10=0$

$\left\{\begin{array}{c}\because 10 \times 10=100 \\ \therefore 100=-25 \times(-4) \\ -29=-25-4\end{array}\right\}$

$\Rightarrow 10 r^{2}-25 r-4 r+10=0$

$\Rightarrow 5 r(2 r-5)-2(2 r-5)=0$

$\Rightarrow(2 r-5)(5 r-2)=0$

Either 2r-5=0, then $r=\frac{5}{2}$

or 5r-2=0, then $r=\frac{2}{5}$

Hence $r=\frac{5}{2}$ or $\frac{2}{5}$

and terms will be if $r=\frac{5}{2}$

$1, \frac{5}{2}, \frac{25}{4}, \frac{125}{8}, \ldots$

if $r=\frac{2}{5},$ then terins will be

$1, \frac{2}{5}, \frac{4}{25}, \frac{8}{125}, \ldots$

Question 20

Three numbers are in A.P. and their sum is 15. If 1, 4 and 19 are added to these numbers respectively, the resulting numbers are in G.P. Find the numbers.

Sol :

Given: Three numbers are in A.P. and their sum = 15

Let a – d, a, a + d be the three number in A.P.

$\therefore a-d+a+a+d=15 \Rightarrow 3 a=15$

$\Leftrightarrow a=\frac{15}{3}=5$

By adding 1,4,19 in then,

We get

a-d+1, a+4, a+d+19

These are in G.P. $\therefore b^{2}=a \cdot c$

$\therefore(a+4)^{2}=(a-d+1)(a+d+19)$

$\Rightarrow a^{2}+8 a+16=a^{2}+a d+19 a-a d-d^{2}-19 d+a+d+19$

$\Rightarrow a^{2}+8 a+16=a^{2}-d^{2}-18 d+20 a+19$

$\Rightarrow 8 a+16=20 a-18 d-d^{2}+19$

$\Rightarrow 8 a+16-20 a+18 d+d^{2}-19=0$

$\Rightarrow d^{2}+18 d-12 a-3=0$

$\Rightarrow d^{2}+18 d-12 \times 5-3=0$

$\Rightarrow d^{2}+18 d-60-3=0$

$\Rightarrow d^{2}+18 d-63=0$

$\left\{\begin{aligned} \because-63 &=+21 \times-3 \\+& 18=+21-3 \end{aligned}\right\}$

$\Rightarrow d^{2}+21 d-3 d-63=0$

$\Rightarrow d(d+21)-3(d+21)=0$

$\Rightarrow(d+21)(d-3)=0$

Either d+21=0, then d=-21

or d-3=0, then d=+3

If d=3 and a=5, then G.P.

5-3,5,5+3 i.e. 2,5,8

If d=-21, then

5+21,5,5-21

$\Rightarrow 26,5,-16$

Question 21

Three numbers form an increasing G.P. If the middle term is doubled, then the new numbers are in A.P. Find the common ratio of the G.P.

Sol :

Three numbers form an increasing G.P.

Let $\frac{a}{r}$ ,a,ar be three numbers in G.P.

Double the middle term, we get

$\frac{a}{r}, 2 a, a r$ will be in A.P.

If $2(2 a)=\frac{a}{r}+a r$

If $4 a=a\left(\frac{1}{r}+r\right)$

If $4=\frac{1}{r}+r \Rightarrow 4 r=1+r^{2}$

$\Rightarrow r^{2}-4 r+1=0$

$\Rightarrow r=\frac{-b \pm \sqrt{b^{2}-4 a c}}{2 a}$

$=\frac{-(-4) \pm \sqrt{(-4)^{2}-4 \times 1 \times 1}}{2 \times 1}$

$=\frac{4 \pm \sqrt{16-4}}{2}=\frac{4 \pm \sqrt{12}}{2}$

$=\frac{4 \pm 2 \sqrt{3}}{2}=2 \pm \sqrt{3}$

$\therefore r=2 \pm \sqrt{3}$

$\because$ The numbers are increasing.

$\therefore r=2+\sqrt{3}$

Question 22

Three numbers whose sum is 70 are in GP. If each of the extremes is multiplied by 4 and the mean by 5, the numbers will be in A.P. Find the numbers.

Sol :

Three numbers are in G.P.

Let numbers be

$\frac{r}{a}, a, a r$

$\therefore \frac{a}{r}+a+a r=70 $

$\Rightarrow a\left(\frac{1}{r}+1+r\right)=70 \ldots(i)$

By multiplying the extremes by 4 and mean by $5,$ then

$\frac{a}{r} \times 4, a \times 5, a r \times 4$

$\frac{4 a}{r}, 5 a, 4 a r$

But these are in A.P.

$\therefore 2(5 a)=\frac{4 a}{r}+4 a r$

$10 a=4 a\left(\frac{1}{r}+r\right)$

$5=2\left(\frac{1}{r}+r\right)$

$\Rightarrow 5 r=2+2 r^{2}$

$\Rightarrow 2 r^{2}-5 r+2=0$

$\Rightarrow 2 r^{2}-r-4 r+2=0$

$\left\{\begin{array}{l}\because 2 \times 2=4 \\ \therefore 4=-1 \times(-4) \\ -5=-1-4\end{array}\right\}$

$\Rightarrow r(2 r-1)-2(2 r-1)=0$

$\Rightarrow(2 r-1)(r-2)=0$

Either 2r-1=0, then $r=\frac{1}{2}$

or r-2=0, then r=2

From (i)

$a\left(\frac{1}{2}+1+2\right)=70$

$\frac{7}{2} a=70 $

$\Rightarrow a=70 \times \frac{2}{7}=20$

$\therefore$ Numbers are (if $r=2)$

$\frac{20}{2}, 20,20 \times 2 \Rightarrow 10,20,40$

If $r=\frac{1}{2},$ then

$ \frac{20}{\frac{1}{2}}, 20,20 \times \frac{1}{2}$

$ \Rightarrow \frac{20 \times 2}{1}, 20,20 \times \frac{1}{2}$

$\Rightarrow 40,20,10$

Question 23

There are four numbers such that first three of them form an A.P. and the last three form a GP. The sum of the first and third number is 2 and that of second and fourth is 26. What are these numbers?

Sol :

There are 4 numbers, such that

First 3 numbers are in A.P. and

last 3 numbers are in GP.

Sum of first and third numbers = 2

and sum of 2nd and 4th = 26

Let a, b, c and d are numbers such that

a, b, c are in A.P.

$\therefore 2 b=a+c$...(i)

and b, c, d are in G.P. $\therefore c^{2}=b d$...(ii)

Also. a+c=2 

$\Rightarrow 2 b=2 \Rightarrow b=\frac{2}{2}=1$...(iii)

and b+d=26 $\Rightarrow 1+d=26$

$\Rightarrow d=26-1=25$...(iv)

Now, $c^{2}=b d=1 \times 25=25=(5)^{2}$

$\therefore c=5$

$a=2 b-c=2 \times 1-5=2-5=-3$

$\therefore$ Numbers are -3,1,5,25

Question 24

(i) If a, b, c are in A.P. as well in G.P., prove that a = b = c.

(ii) If a, b, c are in A.P as well as in G.P., then find the value of $a^{b-c}+b^{c-a}+c^{a-b}$

Sol :

(i) a, b, c are in A.P. as well as in GP.

To prove: a = b = c

a, b, c are in A.P.

$\therefore 2 b=a+c \Rightarrow b=\frac{a+c}{2}$...(i)

$\therefore a, b, c$ are in G.P. 

$\therefore b^{2}=a c$...(ii)

$\Rightarrow\left(\frac{a+c}{2}\right)^{2}=a c $

$\Rightarrow \frac{(a+c)^{2}}{4}=a c$

$\Rightarrow(a+c)^{2}=4 a c $

$\Rightarrow(a+c)^{2}-4 a c=0$

$\Rightarrow(a-c)^{2}=0 $

$\Rightarrow a-c=0$

$ \Rightarrow a=c$...(iii)

From $(i), 2 b=a+c=a+a=2 a$

$\therefore b=a$...(iv)

From (iii) and (iv)

Hence , a=b=c

(ii) a, b, c are in A.P. as well as in G.P. 

$\therefore 2 b=a+c$

and $b^{2}=a c$

and a=b=c [proved in (i)]

Now, $a^{b-c}+b^{c-a}+c^{a-b}$

Since, a=b=c

$\therefore b-c=0, c-a=0$ and $a-b=0$

$\therefore a^{0}+b^{0}+c^{0}=1+1+1$ $\left(\because x^{0}=1\right)$

=3

Question 25

The terms of a G.P. with first term a and common ratio r are squared. Prove that resulting numbers form a G.P. Find its first term, common ratio and the nth term.

Sol :

In a G.P.,

The first term = a

and common ratio = r

GP. is a, ar, ar²

Squaring we get

$a^{2}, a^{2} r^{2}, a^{2} r^{4}$ are in G.P.

if $b^{2}=4 a c$

$\Rightarrow\left(a^{2} r^{2}\right)^{2}=a^{2} \times a^{2} r^{4} \Rightarrow a^{4} r^{4}=a^{4} r^{4}$

Which is true The first term is $a^{2}$ and common ratio is $r^{2}$ 

The n th term will be $a_{n}=a r^{n-1}=a^{2}\left(r^{n-1}\right)^{2}=a^{2} r^{2 n-2}$

Question 26

Show that the products of the corresponding terms of two G.P.’s a, ar, ar², …, a r^{n-1} \text { and } A, A R, A R^{2}, \ldots, A R^{n-1} form a G.P. and find the common ratio.

Sol :

It has to be proved that the sequence

$\mathrm{aA}, \operatorname{arAR}, \mathrm{ar}^{2} \mathrm{AR}^{2}, \ldots, \mathrm{ar}^{n-1} \mathrm{AR}^{n-1}$ and forms a G.P.

Hence, $\frac{\text { Second term }}{\text { First term }}=\frac{a r \mathrm{AR}}{a \mathrm{~A}}=r \mathrm{R}$

and $\frac{\text { Third term }}{\text { Second term }}=\frac{a r^{2} A R^{2}}{a r A R}=r R$

Thus, the above sequence forms a G.P. and the common ratio is rR

Question 27

(i) If a, b, c are in G.P. show that $\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ are also in G.P.

(ii) If K is any positive real number and $K^{a}, K^{b} K^{c}$ are three consecutive terms of a G.P., prove that a, b, c are three consecutive terms of an A.P.

(iii) If p, q, r are in A.P., show that pth, qth and rth terms of any G.P. are themselves in GP.

Sol :

(i) a, b, c are in G.P.

$\therefore b^{2}=a c$

$\frac{1}{a}, \frac{1}{b}, \frac{1}{c}$ will be in G.P.

if $\left(\frac{1}{b}\right)^{2}=\frac{1}{a} \times \frac{1}{c} \Rightarrow \frac{1}{b^{2}}=\frac{1}{a c}$

$\Rightarrow a c=b^{2}$ (By cross multiplication)

which is given

Hence proved

(ii) $\mathrm{K}$ is any positive number $\mathrm{K}^{a}, \mathrm{~K}^{b}, \mathrm{~K}^{c}$ are in G.P.

then $\left(\mathrm{K}^{b}\right)^{2}=\mathrm{K}^{a} \times \mathrm{K}^{c}$

$\Rightarrow \mathrm{K}^{2 b}=\mathrm{K}^{a+c}$

$\Rightarrow 2 b=a+c$

Hence, a, b, c are in A.P.

(iii) $p, q, r$ are in A.P. $\therefore 2 q=p+r$

p th term in $\mathrm{G.P.}=\mathrm{AR}^{p-1}$

q th term $=\mathrm{AR}^{q-1}$

rth term $=\mathrm{AR}^{r-1}$

These will be in G.P. if $\left(\mathrm{AR}^{q-1}\right)^{2}=\mathrm{AR}^{p-1} \times \mathrm{AR}^{r-1}$

if $\mathrm{A}^{2} \mathrm{R}^{2 q-2}=\mathrm{A}^{2} \mathrm{R}^{p-1+r-1}$

if $A^{2} R^{2 q-2}=A^{2} R^{p+r-2}$

if $R^{2 q-2}=R^{p+r-2}$

Comparing, we get $2 q-2=p+r^{\bullet}-2$

$\Rightarrow 2 q=p+r$

$\Rightarrow p, q, r$ are in A.P. which is given

Hence proved

Question 28

If a, b, c are in GP., prove that the following are also in G.P.

(i) $a^{3}, b^{3}, c^{3}$

(ii) $a^{2}+b^{2}, a b+b c, b^{2}+c^{2}$

Sol :

(i) a, b, c are in G.P.

$\therefore b^{2}=a c$

$a^{3}, b^{3}, c^{3}$ are in G.P.

if $\left(b^{3}\right)^{2}=a^{3} \times c^{3}$

if $\left(b^{2}\right)^{3}=(a \times c)^{3}$

if $b^{2}=a c$

which is given

Hence proved

(ii) $a^{2}+b^{2}, a b+b c, b^{2}+c^{2}$ will be in G.P. $a^{2}+b^{2}, a b+b c, b^{2}+c^{2}$ are in G.P.

If $\frac{a b+b c}{a^{2}+b^{2}}=\frac{b^{2}+c^{2}}{a b+b c}$

i.e., if $\frac{a(a r)+\operatorname{ar}\left(a r^{2}\right)}{a^{2}+(a r)^{2}}=\frac{(a r)^{2}+\left(a r^{2}\right)^{2}}{a(a r)+a r\left(a r^{2}\right)}$

$\Rightarrow$ If $\frac{a^{2} r\left(1+r^{2}\right)}{a^{2}\left(1+r^{2}\right)}=\frac{a^{2} r^{2}\left(1+r^{2}\right)}{a^{2} r\left(1+r^{2}\right)}$

$\Rightarrow$ If r=r which is true. $\therefore a^{2}+b^{2}, a b+b c, b^{2}+c^{2}$ are in G.P.

Question 29

If a, b, c, d are in G.P., show that

(i) $a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+d^{2}$ are in G.P.

(ii) $(b-c)^{2}+(c-a)^{2}+(d-b)^{2}=(a-d)^{2}$

Sol :

a, b, c, d are in G.P.

Let r be the common ratio, then a = a

$b=a r, c=a r^{2}, d=a r^{3}$

(i) $a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+d^{2}$ are in G.P

$\therefore a^{2}+b^{2}=a^{2}+a^{2} r^{2}=a^{2}\left(1+r^{2}\right)$

$b^{2}+c^{2}=a^{2} r^{2}+a^{2} r^{4}=a^{2} r^{2}\left(1+r^{2}\right)$

$c^{2}+d^{2}=a^{2} r^{4}+a^{2} r^{6}=a^{2} r^{4}\left(1+r^{2}\right)$

$a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+d^{2}$ will be in G.P.

if $\left(b^{2}+c^{2}\right)^{2}=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)$

Now, $\left(b^{2}+c^{2}\right)^{2}=\left[a^{2} r^{2}\left(1+r^{2}\right)\right]^{2}$

$=a^{4} r^{4}\left(1+r^{2}\right)^{2}$...(i)

$\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)=\left[a^{2}\left(1+r^{2}\right)\right]\left[a^{2} r^{4}\left(1+r^{2}\right)\right]^{2}$

$=a^{4} r^{4}\left(1+r^{2}\right)^{2}$...(ii)

From (i) and (ii)

$\left(b^{2}+c^{2}\right)^{2}=\left(a^{2}+b^{2}\right)\left(c^{2}+d^{2}\right)$

Hence, $a^{2}+b^{2}, b^{2}+c^{2}, c^{2}+d^{2}$ are in G.P.

(ii) Show that

$(b-c)^{2}+(c-a)^{2}+(d-b)^{2}=(a-d)^{2}$

L.H.S. $=(b-c)^{2}+(c-a)^{2}+(d-b)^{2}$

$=\left(a r-a r^{2}\right)^{2}+\left(a r^{2}-a\right)^{2}+\left(a r^{3}-a r\right)^{2}$

$=a^{2} r^{2}(1-r)^{2}+a^{2}\left(r^{2}-1\right)^{2}+a^{2} r^{2}\left(r^{2}-1\right)^{2}$

$=a^{2}\left[r^{2}\left(1-r^{2}-2 a r\right)+r^{4}-2 r^{2}+1+r^{2}\left(r^{4}-\right.\right.\left.\left.2 r^{2}+1\right)\right]$

$=a^{2}\left[r^{2}-r^{4}-2 a r^{3}+r^{4}-2 r^{2}+1+r^{6}-2 r^{4}\right.\left.\left.+r^{2}\right)\right]$

$=a^{2}\left(r^{6}-2 r^{3}+1\right)$

R.H.S. $=(a-d)^{2}=\left(a-a r^{3}\right)^{2}=a^{2}\left(1-r^{3}\right)^{2}$

$=a^{2}\left(1+r^{6}-2 r^{2}\right)$

$=a^{2}\left[r^{6}-2 r^{2}+1\right]$

∴L.H.S=R.H.S

Question 30

The number of bacteria in a certain culture doubles every hour. If there were 30 bacteria present in the culture originally, how many bacteria will be present at the end of 2nd hour, 4th hour and nth hour?

Sol :

Bacteria in the beginning = 30 = a

After every hours, it doubles itself

After 1 hour it becomes = 30 x 2 = 60 = $a^r$

After 2 hours it will becomes = 60 x 2 = 120 = $a^2$

After 3 hours, it will becomes = 120 x 2 = 240 = $a^3$

After 4 hours it will becomes = 240 x 2 = 480 = $a^4$

∴ After n hour, it will become = $ar^n$

Question 31

The length of the sides of a triangle form a G.P. If the perimeter of the triangle is 37 cm and the shortest side is of length 9 cm, find the lengths of the other two sides.

Sol :

Lengths of a triangle are in GP. and its sum is 37 cm

Let sides be $a, a r, a r^{2}$ 

$a+a r+a r^{2}=37$

a=9

$\therefore 9+9 r+9 r^{2}=37 $

$\Rightarrow 9 r+9 r^{2}=37-9=28$

$9 r^{2}+9 r-28=0$

$ \Rightarrow 9 r^{2}+21 r-12 r=28=0$

$\left\{\begin{array}{l}\because 28 \times 9=252 \\ \therefore 252=21 \times-12 \\ 9=21-12\end{array}\right\}$

$\Rightarrow 3 r(3 r+7)-4(3 r+7)=0$

(3r+7)(3r-4)=0

3r+7=0 then  3r=-7, 

$r=\frac{-7}{3}$

or (3r-4)=0 then 3r=4$\Rightarrow r=\frac{4}{3}$

Sides are $9,9 \times \frac{4}{3}, 9 \times \frac{4}{3} \times \frac{4}{3}$

=9,12,16 cm

ML Aggarwal Solution Class 10 Chapter 9 Arithmetic and Geometric Progressions Exercise 9.3

Q1 Q2 Q3 Q4 Q5 Q6 Q7 Q8 Q9 Q10 Q11 Q12 Q13 Q14 Q15 Q16 Q17 Q18 Q19 Q20 Q21 Q22 Q23 Q24 Q25

 Exercise 9.3

Question 1

Find the sum of the following A.P.s :

(i) 2, 7, 12, … to 10 terms

(ii) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$ to 11 terms

Sol :

(i) 2, 7, 12, … to 10 terms

Here a = 2, d = 7 – 2 = 5 and n = 10

$\mathrm{S}_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 2+(10-1) \times 5]$

$=5(4+45)=5 \times 49=245$

(ii) $\frac{1}{15}, \frac{1}{12}, \frac{1}{10}, \ldots$ to 11 terms

$a=\frac{1}{15}$

$d=\frac{1}{12}-\frac{1}{15}$

$=\frac{5-4}{60}=\frac{1}{60}$ or $\frac{1}{10}-\frac{1}{12}$

$=\frac{6-5}{60}=\frac{1}{60}$

$n=11$

$\therefore \mathrm{S}_{11}=\frac{n}{2} \times[2 a+(n-1) d]$

$=\frac{11}{2} \times\left[2 \times \frac{1}{15}+(11-1) \times \frac{1}{60}\right]$

$=\frac{11}{2} \times\left[\frac{2}{15}+\frac{1}{6}\right]=\frac{11}{2} \times\left[\frac{4+5}{30}\right]$

$=\frac{11}{2} \times \frac{9}{30}=\frac{33}{20}$ or

$=1 \frac{13}{20}$

Question 2

How many terms of the A.P. 27, 24, 21, …, should be taken so that their sum is zero?

Sol :

A.P. = 27, 24, 21,…

a = 27

d = 24 – 27 = -3

$S_{n}=0$

Let n terms be there in A.P.

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 0=\frac{n}{2}[(2 \times 27)+(n-1)(-3)]$

$\Rightarrow 0=n[54-3 n+3]$

$\Rightarrow n[57-3 n]=0$

$\Rightarrow(57-3 n)=\frac{0}{n}=0$

$\Rightarrow 3 n=57$

$\therefore n=\frac{57}{3}=19$

Question 3

Find the sums given below :

(i) 34 + 32 + 30 + … + 10

(ii) – 5 + ( – 8) + ( – 11) + … + ( – 230)

Sol :

(i) 34 + 32 + 30 + … + 10

Here, a = 34, d = 32 – 34 = -2, l = 10

Tn = a + (n – 1)d

10 = 34 + (n – 1)(-2)

-24 = -2 (n – 1)

$=\frac{-24}{-2}=n-1 \Rightarrow n-1=12$

$\therefore n=12+1=13$

$\mathrm{~S}_{n}=\frac{n}{2}[a+l]$

$=\frac{13}{2}[34+10]=\frac{13}{2} \times 44=286$

(ii) -5+(-8)+(-11)+....+(-230)

Here, a=-5, d=-8-(-5)=-8+5=-3

l=-230

$\therefore l=a+(n-1) d$

$ \Rightarrow-230=-5+(n-1)(-3)$

$-230+5=-3(n-1)$

$ \Rightarrow-225=-3(n-1)$

$\frac{-225}{-3}=n-1$

$ \Rightarrow n-1=75$

$\Rightarrow n=75+1=76$

$\therefore \mathrm{S}_{n}=\frac{n}{2}[a+l]$

$=\frac{76}{2}[-5+(-230)]$

$=38[-5-230]=38 \times(-235)=-8930$

Question 4

In an A.P. (with usual notations) :

(i) given a = 5, d = 3, a_n = 50, find n and S_n

(ii) given a = 7, a₁₃ = 35, find d and S₁₃

(iii) given d = 5, S₉ = 75, find a and a₉

(iv) given a = 8, a_n = 62, S_n = 210, find n and d

(v) given a = 3, n = 8, S = 192, find d.

Sol :

(i) a = 5, d = 3, an = 50

an = a + (n – 1 )d

50 = 5 + (n – 1) x 3

⇒ 50 – 5 = 3(n – 1)

$\Rightarrow 45=3(n-1)$

$ \Rightarrow \frac{45}{3}=n-1$

$\Rightarrow n-1=15$

$ \Rightarrow n=15+1=16$

$\therefore n=16$

and $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ 

$=\frac{16}{2}[2 \times 5+(16-1) \times 3]=8[10+45]$

$=8 \times 55=440$

(ii) $a=7, a_{13}=35$

$a_{n}=a+(n-1) d$

$35=7+(13-1) d $

$\Rightarrow 35-7=12 d$

$\Rightarrow 28=12 d $

$\Rightarrow d=\frac{28}{12}=\frac{7}{3}=2 \frac{1}{3}$

and $\mathrm{S}_{13}=\frac{n}{2} \cdot[2 a+(n-1) d]$

$=\frac{13}{2}\left[2 \times 7+(13-1) \times \frac{7}{3}\right]$

$=\frac{13}{2}[14+28]=\frac{13}{2} \times(42)$

$=13 \times 21=273$

(iii) $d=5, \mathrm{~S}_{9}=75$

$a_{n}=a+(n-1) d$

$a_{9}=a+(9-1) \times 5$

=a+40...(i)

$\mathrm{S}_{9}=\frac{n}{2}[2 a+(n-1) d]$

$75=\frac{9}{2}[2 a+8 \times 5]$

$\frac{150}{9}=2 a+40$

$2 a=\frac{150}{9}-40=\frac{50}{3}-40$

$2 a=\frac{-70}{3} \Rightarrow a=\frac{-70}{2 \times 3}$

$a=\frac{-35}{3}$

From (i)

$a_{9}=a+40=\frac{-35}{3}+40$

$=\frac{-35+120}{3}=\frac{85}{3}$

$\therefore a=\frac{-35}{3}, a_{9}=\frac{85}{3}$

(iv) $a=8, a_{n}=62, \mathrm{~S}_{n}=210$

$a_{n}=a+(n-1) d$

$62=8+(n-1) d$

$(n-1) d=62-8=54$...(i)

$\mathrm{~S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$210=\frac{n}{2}[2 \times 8+54]$ [From (i)]

$420=n(16+54) \Rightarrow 420=70 n$

$n=\frac{420}{70}=6$

$\therefore(6-1) d=54$

$\Rightarrow 5 d=54$

$\Rightarrow d=\frac{54}{5}$

Hence $d=\frac{54}{5}$ and $n=6$

(v) $a=3, n=8, \mathrm{~S}=192$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$192=\frac{8}{2}[2 \times 3+7 \times d]$

192=4[6+7d]

$ \Rightarrow \frac{192}{4}=6+7 d$

$\Rightarrow 48=6+7 d $

$\Rightarrow 7 d=48-6=42$

$d=\frac{42}{7}=6$

$\therefore d=6$

Question 5

(i) The first term of an A.P. is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

(ii) The sum of first 15 terms of an A.P. is 750 and its first term is 15. Find its 20th term.

Sol :

(i) First term of an A.P. (a) = 5

Last term (l) = 45

Sum = 400

l = a + (n – 1 )d

45 = 5 + (n – 1)d

⇒ (n – 1)d = 45 – 5 = 40 …(i)

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$400=\frac{n}{2}[2 \times 5+40]$

$ \Rightarrow 800=n(10+40)$

50n=800

$\Rightarrow n=\frac{800}{50}=16$

From (i)

(16-1) d=40 

$\Rightarrow 15 d=40$

$ \Rightarrow d=\frac{40}{15}$

$\therefore d=\frac{8}{3}$ and $n=16$

(ii) Let a be the first term and d be the common difference.

Now, a=15

Sum of first n terms of an AP is given by,

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow \mathrm{S}_{15}=\frac{15}{2}[2 a+(15-1) d]$

$\Rightarrow 750=\frac{15}{2}(2 a+14 d)$

$\Rightarrow a+7 d=50$

$\Rightarrow 15+7 d=50$

$\Rightarrow 7 d=35$

$\Rightarrow d=5$

Now, 20 th term $=a_{20}=a+19 d$

$=15+19 \times 5=15+95=110$

Question 6

The first and the last terms of an A.P. are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Sol :

First term of an A.P. (a) = 17

and last term (l) = 350

d= 9

$1=T_{n}=a+(n-1) d$

$350=17+(n-1) \times 9$

$\Rightarrow 350-17=9(n-1)$

$\Rightarrow 333=9(n-1) $

$\Rightarrow n-1=\frac{333}{9}=37$

n=37+1=38

and $\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{38}{2}[2 \times 17+(38-1) \times 9]$

$=19[34+37 \times 9]=19[34+333]$

$=19 \times 367=6973$

Hence n=38 and $\mathrm{S}_{n}=6973$

Question 7

Solve for x : 1 + 4 + 7 + 10 + … + x = 287.

Sol :

1 + 4 + 7 + 10 + .. . + x = 287

Here, a = 1, d = 4 – 1 = 3, n = x

l = x = a = (n – 1)d = 1 + (n – 1) x 3

$\Rightarrow x-1=(n-1) d$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$287=\frac{n}{2}[2 \times 1+(n-1) 3]$

574=n(2-3n-3)

$\Rightarrow 3 n^{2}-n-574=0$

$\Rightarrow 3 n^{2}-42 n+41 n-574=0$

$\Rightarrow 3 n(n-14)+41(n-14)=0$

$\Rightarrow(n-14)(3 n+41)=0$

Either n-14=0, then n=14

or 3n+41=0, then 3n=-41 

$\Rightarrow n=\frac{-41}{3}$

which is not possible being negative.

$\therefore n=14$

Now, $x=a+(n-1) d$

$=1+(14-1) \times 3=1+13 \times 3$

$=1+39=40$

$\therefore x=40$

Question 8

(i) How many terms of the A.P. 25, 22, 19, … are needed to give the sum 116 ? Also find the last term.

(ii) How many terms of the A.P. 24, 21, 18, … must be taken so that the sum is 78 ? Explain the double answer.

Sol :

(i) A.P. is 25, 22, 19, …

Sum = 116

Here, a = 25, d = 22 – 25 = -3

Let number of terms be n, then

$116=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 232=n[2 \times 25+(n-1)(-3)]$

$\Rightarrow 232=n[50-3 n+3] \Rightarrow 232=n(53-3 n)$

$\Rightarrow 232=53 n-3 n^{2}$

$\Rightarrow 3 n^{2}-53 n+232=0$

$\left\{\begin{array}{l}\because 232 \times 3=696 \\ \therefore 696=-24 \times(-29) \\ -53=-24-29\end{array}\right\}$

$\Rightarrow 3 n^{2}-24 n-29 n+232=0$

$\Rightarrow 3 n(n-8)-29(n-8)=0$

$\Rightarrow(n-8)(3 n-29)=0$

Either n-8=0, then n=8

or 3n-29=0, then 3n=29 

$\Rightarrow n=\frac{29}{3}$

which is not possible because of fractioin

$\therefore n=8$

Now, $\mathrm{T}=a+(n-1) d$

$=25+7 \times(-3)=25-21=4$

(ii) A.P. is 24,21,18, ...

Sum =78

Here, a=24, d=21-24=-3

$\mathrm{~S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 78=\frac{n}{2}[2 \times 24+(n-1)(-3)]$

$\Rightarrow 156=n(48-3 n+3)$

$\Rightarrow 156=51 n-3 n^{2}$

$\Rightarrow 3 n^{2}-51 n+156=0$

$\Rightarrow 3 n^{2}-12 n-39 n+156=0$

$\left\{\begin{array}{c}\because 156 \times 3=468 \\ \therefore 468=-12 \times-39 \\ -51=-12-39\end{array}\right\}$

$\Rightarrow 3 n(n-4)-39(n-4)=0$

$\Rightarrow(n-4)(3 n-39)=0$

Either n-4=0, then n=4

or 3n-39=0, then 3n=39 

$\Rightarrow n=13$

$\therefore n=4$ and 13

$n_{4}=a+(n-1) d=24+3(-3)$

=24-9=15

$n_{13}=24+12(-3)=24-36=-12$

$\therefore$ Sum of 5 th term to 13 term $=0$

$(\because 12+9+6+3+0+(-3)+(-6)+(-9)+(-12)=0$

Question 9

Find the sum of first 22 terms, of an A.P. in which d = 7 and $a_{22}$ is 149.

Sol :

Sum of first 22 terms of an A.P. whose d = 7

$a_{22}=149$ and n=22

149=a+(n-1)d$=a+21 \times 7$

$149=a+147 \Rightarrow a=149-147=2$

$\therefore \mathrm{S}_{22}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{22}{2}[2 \times 2+(22-1)(7)]$

$=11[4+21 \times 7]=11 \times[4+147]$

$=11 \times 151=1661$

Question 10

(i) Find the sum of first 51 terms of the A.P. whose second and third terms are 14 and 18 respectively.

(ii) If the third term of an A.P. is 1 and 6th term is – 11, find the sum of its first 32 terms.

Sol :

Sum of first 51 terms of an A.P. in which

$T_{2}=14, T_{3}=18$

$\therefore d=T_{3}-T_{2}=18-14=4$

and $a=T_{1}=14-4=10, n=51$

Now, $\mathrm{S}_{51}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{51}{2}[2 \times 10+(51-1) \times 4]$

$=\frac{51}{2}[20+50 \times 4]$

$=\frac{51}{2}[20+200]$

$=\frac{51}{2} \times 220=5610$

(ii) $\mathrm{T}_{3}=1, \mathrm{~T}_{6}=-11, n=32$

$a+2 d=1$..(i)

$a+5 d=-11$..(ii)

Subtracting (i) and (ii),

-3d=12 

$\Rightarrow d=\frac{12}{-3}=-4$

Substitute the value of d in eq. (i)

$a+2(-4)=1 \Rightarrow a-8=1$

a=1+8=9

$\therefore a=9, d=-4$

$\mathrm{S}_{32}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{32}{2}[2 \times 9+(32-1) \times(-4)]$

$=16[18+31 \times(-4)]$

$=16[18-124]=16 \times(-106)]=-1696$

Question 11

If the sum of first 6 terms of an A.P. is 36 and that of the first 16 terms is 256, find the sum of first 10 terms.

Sol :

$S_{6}=36$

$S_{16}=256$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\therefore \mathrm{S}_{6}=\frac{6}{2}[2 a+(6-1) d]=36$

$\Rightarrow 3[2 a+5 d]=36$

$\Rightarrow 2 a+5 d=12$...(i)

and $S_{16}=\frac{16}{2}[2 a+(16-1) d]=256$

$8[2 a+15 d]=256$

$2 a+15 d=32$...(ii)

Subtracting (i) from (ii)

-10d=-20

$\Rightarrow d=\frac{-20}{-10}=2$

Substitute the value of d in eq. (i)

2a+5d=12 

$\Rightarrow 2 a+5 \times 2=12$

$\Rightarrow 2 a+10=12 $

$\Rightarrow 2 a=12-10=2$

Now, $S_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 1+(10-1) \times 2]$

$=5[2+9 \times 2]$

=5[2+18]

$=5 \times 20=100$

Question 12

Show that $a_{1}, a_{2}, a_{3}, \ldots$ form an A.P. where $\mathbf{a}_{\mathbf{n}}$ is defined as $a_{n}=3+4 n$ Also find the sum of first 15 terms.

Sol :

$a_{n}=3+4 n$

$a_{1}=3+4 \times 1=3+4=7$

$a_{2}=3+4 \times 2=3+8=11$

$a_{3}=3+4 \times 3=3+12=15$

$a_{4}=3+4 \times 4=3+16=19$

and so on Here, a = 1 and d = 11 – 7 = 4

$\mathrm{S}_{15}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{15}{2}[2 \times 7+(15-1) \times 4]$

$=\frac{15}{2}[14+14 \times 4]$

$=\frac{15}{2}[14+56]$

$=\frac{15}{2} \times 70=525$

Question 13

(i)If $a_{n}=3-4 n $show that $a_{1}, a_{2}, a_{3}, \ldots$ form an A.P. Also find $S_{20}$.

(ii) Find the common difference of an A.P. whose first term is 5 and the sum of first four terms is half the sum of next four terms.

Sol :

(i) $a_n$ = 3 – 4n

$a_1$ = 3 – 4 x 1 = 3 – 4 = -1

$a_2$ = 3 – 4 x 2 = 3 – 8 = -5

$a_3$ = 3 – 4 x 3 = 3 – 12 = -9

$a_4$ = 3 – 4 x 4 = 3 – 16 = -13 and so on

Here, a = -1, d = -5 – ( -1) = -5 + 1 = -4

$\mathrm{Now}, \mathrm{S}_{20}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{20}{2}[2 \times(-1)+(20-1) \times(-4)]$

$=10[-2+19 \times(-4)]$

$=10[-2-76]=10 \times(-78)=-780$

(ii) Let a and d be the first term and common difference of A.P. respectively. Given, a=5

difference of A.P. respectively. 

Given, a=5

$a_{1}+a_{2}+a_{3}+a_{4}=\frac{1}{2}\left(a_{5}+a_{6}+a_{7}+a_{8}\right)$

$\therefore a+(a+d)+(a+2 d)+(a+3 d)=\frac{1}{2}[(a+4 d)+(a+5 d)+(a+6 d)+(a+7 d)]$

$\Rightarrow 2(4 a+6 d)=(4 a+22 d)$

$\Rightarrow 2(20+6 d)=(20+22 d)$ $(\because a=5)$

$\Rightarrow 40+12 d=20+22 d$

$\Rightarrow 10 d=20$

$\Rightarrow d=2$

Thus, the common difference of A.P. is 2 .

Question 14

The sum of first n terms of an A.P. whose first term is 8 and the common difference is 20 equal to the sum of first 2n terms of another A.P. whose first term is – 30 and the common difference is 8. Find n.

Sol :

In an A.P.

$S_{n}=S_{2 n}$

For the first A.P. a = 8, d = 20

and for second A.P. a = -30, d = 8

$\mathrm{Now}, \mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{n}{2}[2 \times 8+(n-1) \times 20]$

$=\frac{n}{2}[16+20 n-20]=\frac{n}{2}[20 n-4]$

$=10 n^{2}-2 n$...(i)

Similarly,

$\mathrm{S}_{2 n}=2 \frac{n}{2}[2 a+(2 n-1) d]$

$=n[2 \times(-30)+(2 n-1) \times 8]$

=n[-60+16 n-8]

=n(16 n-68)

$=16 n^{2}-68 n$

$\because \mathrm{S}_{n}=2_{2 n}$

$\therefore 10 n^{2}-2 n=16 n^{2}-68 n$

$\Rightarrow 16 n^{2}-10 n^{2}-68 n+2 n=0$

$\Rightarrow 6 n^{2}-66 n=0$

$ \Rightarrow n^{2}-11 n^{\doteq}=0$

n(n-11)=0

Either n=0 which is not possible

or n-11=0, then n=11

$\therefore n=11$

Question 15

The sum of first six terms of an arithmetic progression is 42. The ratio of the 10th term to the 30th term is $\frac{1}{3}$. Calculate the first and the thirteenth term.

Sol :

$\mathrm{T}_{10}: \mathrm{T}_{30}=1: 3, \mathrm{~S}_{6}=42$

Let a be the first term and d be a common difference, then

$\frac{a+9 d}{a+29 d}=\frac{1}{3} \Rightarrow 3 a+27 d=a+29 d$

$\Rightarrow 3 a-a=29 d-27 d$

$\Rightarrow 2 a=2 d \Rightarrow a=d$

Now, $S_{6}=42=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 42=\frac{6}{2}[2 a+(6-1) d]$

$\Rightarrow 42=3[2 a+5 d]$

$\Rightarrow 14=2 a+5 d $

$\Rightarrow 14=2 a+5 a $ $(\because d=a)$

$\Rightarrow 7 a=14 $

$\Rightarrow a=\frac{14}{7}=2$

$\therefore a=d=2$

Now, $T_{13}=a+(n-1) d$

$=2+(13-1) \times 2=2+12 \times 2$

=2+24=26

$\therefore$ 1st term is 2 and thirteenth term is 26

Question 16

In an A.P., the sum of its first n terms is 6n – n². Find is 25th term.

Sol :

$S_{n}=6 n-n^{2}$

$T_{25}=?$

$\mathrm{S}_{(n-1)}=6(n-1)-(n-1)^{2}$

$=6 n-6-\left(n^{2}-2 n+1\right)$

$=6 n-6-n^{2}+2 n-1=8 n-n^{2}-7$

$a_{n}=\mathrm{S}_{n}-\mathrm{S}_{n-1}$

$=6 n-n^{2}-8 n+n^{2}+7$

=-2 n+7

$a_{25}=-2(25)+7=-50+7=-43$

Question 17

If the sum of first n terms of an A.P. is 4n – n², what is the first term (i. e. $S_1$)? What is the sum of the first two terms? What is the second term? Also, find the 3rd term, the 10th term, and the nth terms?

Sol :

$S_{n}=4 n-n^{2}$

$S_{n}-1=4(n-1)-(n-1)^{2}$

$=4 n-4-\left(n^{2}-2 n+1\right)$

$=4 n-4-n^{2}+2 n-1=6 n-n^{2}-5$

$\therefore a_{n}=\mathrm{S}_{n}-\mathrm{S}_{n-1}=4 n-n^{2}-\left(6 n-n^{2}-5\right)$

$=4 n-n^{2}-6 n+n^{2}+5$

=-2 n+5

$a_{1}=-2 \times 1+5=-2+5=3$

$a_{2}=-2 \times 2+5=-4+5=1$

$a_{3}=-2 \times 3+5=-6+5=-1$

$a_{4}=-2 \times 4+5=-8+5=-3$

$a_{10}=-2 \times 10+5=-20+5=-15$

$\mathrm{S}_{2}=4 n-n^{2}=4 \times 2-(2)^{2}$

=8-4=4

Hence, $\mathrm{S}_{2}=4, a_{1}=1, a_{3}=-1, a_{10} \mid=-15$

$a_{n}=-2 n+5$

Question 18

If $S_{n}$ denotes the sum of first n terms of an A.P., prove that $S_{30} = 3(S_{20} – S_{10})$.

Sol :

Sn denotes the sum of first n terms of an A.P.

To prove: $S_{30} = 3(S_{20} – S_{10})$

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$\therefore \mathrm{S}_{10}=\frac{10}{2}[2 a+(10-1) d]=5(2 a+9 d)$

=10 a+45d

$\mathrm{S}_{20}=\frac{20}{2}[2 a+(20-1) d]=10(2 a+19 d)$

=20a+190d

$\mathrm{S}_{30}=\frac{30}{2}[2 a+(30-1) d=15(2 a+29 d)$

=30a+435d

Now, $\mathrm{R.H.S.}=3\left(\mathrm{~S}_{20}-\mathrm{S}_{10}\right)$

$=3[20 a+190 d-10 a-45 d]$

$=3[10 a+145 d]$

$=30 a+435 d$

$=\mathrm{S}_{30}=\mathrm{L} . \mathrm{H.S}$

Question 19

(i) Find the sum of first 1000 positive integers.

(ii) Find the sum of first 15 multiples of 8.

Sol :

(i) Sum of first 1000 positive integers

i. e., 1 + 2 + 3+ 4 + … + 1000

Here, a = 1, d = 1, n = 1000

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{1000}{2}[2 \times 1+(1000-1) 1]$

=500[2+999]

$=500 \times 1001$

=500500

(ii) Sum of first 15 multiples of 8

8+16+24+32+...120

Here, a=8, d=8, n=15

$\therefore \mathrm{S}_{15}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{15}{2}[2 \times 8+(15-1) \times 8]$

$=\frac{15}{2}[16+14 \times 8]$

$=\frac{15}{2}[16+112]$

$=\frac{15}{2} \times 128=15 \times 64=960$

Question 20

(i) Find the sum of all two digit natural numbers which are divisible by 4.

(ii) Find the sum of all natural numbers between 100 and 200 which are divisible by 4.

(iii) Find the sum of all multiples of 9 lying between 300 and 700.

(iv) Find the sum of all natural numbers less than 100 which are divisible by 6.

Sol :

(i) Sum of two digit natural numbers which are divisible by 4

which are 12, 16, 20, 24, …, 96

Here, a = 12, d = 16 – 12 = 4, l = 96

$\therefore l=96=a+(n-1) d=12+(n-1) \times 4$

96=12+4n-4=8+4n

$\Rightarrow 4 n=96-8=88$

$ \Rightarrow n=\frac{88}{4}=22$

$\therefore \mathrm{S}_{22}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{22}{2}[2 \times 12+(22-1) \times 4]$

$=11[24+21 \times 4]=11[24+84]$

$=11 \times 108=1188$

(ii) Sum of all natural numbers between 100 and 200 which are divisible by 4 which are

104,108,112,116, ...196

Here, $a=104, d=108-104=4, l=196$

$l=a_{n}=196=a+(n-1) d$

$\Rightarrow 196=104+(n-1) \times 4$

196-104=(n-1) 4

$ \Rightarrow 92=(n-1) 4$

$n-1=\frac{92}{4}=23$

$\therefore n=23+1=24$

$\mathrm{Now}, \mathrm{S}_{24}=\frac{n}{2}[2 \dot{a}+(n-1) d]$

$=\frac{24}{2}[2 \times 104+(24-1) \times 4]$

$=12[208+23 \times 4]$

$=12 \times[208+92]$

$=12 \times 300=3600$

(iii) Sum of all natural numbers multiple of 9 lying between 300 and 700 which are

306,315,324,333, .... 693

Here, a=306, d=9, l=693

$l=a_{n}=693=a+(n-1) d$

$=306+(n-1) \times 9$

693-306=9(n-1)

387=9(n-1)

$\Rightarrow n-1=\frac{387}{9}=43$

$\therefore n=43+1=44$

$\mathrm{S}_{44}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{44}{2}[2 \times 306+(44-1) \times 9]$

$=22[6.12+43 \times 9]$

$=22[612+387]=22 \times 999=21978$

$\begin{array}{r}999 \\\times 22 \\\hline 1998 \\1998 \times \\\hline 21978 \\\hline\end{array}$

(iii) Sum of all natural numbers less then 100 which are divisible by 6 which are

6,12,18,24, ..., 96

Here, a=6, d=6, l=96

$a_{n}$=l=96=a+(n-1) d

$\Rightarrow 96=6+(n-1) \times 6$

$96-6=6(n-1)$

$\frac{90}{6}=n-1$

$ \Rightarrow n-1=15$

n=15+1=16

$\therefore \mathrm{S}_{16}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{16}{2}[2 \times 6+(16-1) \times 6]$

$=8[12+15 \times 6]=8[12+90]$

$=8 \times 102=816$

Question 21

(i) Find the sum of all two digit odd positive numbers.

(ii) Find the sum of all 3-digit natural numbers which are divisible by 7.

(iii) Find the sum of all two digit numbers which when divided by 7 yield 1 as the

Sol :

(i) Sum of all two-digit odd positive numbers which are 11, 13, 15, …, 99

Here, a = 11, d = 2, l = 99

$a_{n}=l=a+(n-1) d$

$99=11+(n-1) \times 2 $

$\Rightarrow 99-11=2(n-1)$

$\Rightarrow 88=2(n-1)$

$ \Rightarrow n-1=\frac{88}{2}=44$

$\therefore n=44+1=45$

$\mathrm{S}_{45}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{45}{2}[2 \times 11+(45-1) \times 2]$

$=\frac{45}{2}[22+44 \times 2]$

$=\frac{45}{2}[22+88]$

$=\frac{45}{2} \times 110=2475$

(ii) Sum of all 3-digit natural numbers which are divisible by 7 which are 105,112,119 ... 994

Here, a=105, d=112-105=7, l=994

$\therefore l=T_{n}=994=a+(n-1) d$

$\Rightarrow 994=105+(n-1) \times 7$

$\Rightarrow 994-105=(n-1) 7$

$ \Rightarrow 889=7(n-1)$

$\frac{889}{7}=n-1 $

$\Rightarrow n-1=127$

$\Rightarrow n=127+1=128$

$\therefore \mathrm{S}_{128}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{128}{2}[2 \times 105+(128-1) \times 7]$

$=64[210+889]=64 \times 1099=70336$

(iii) The 2 -digit number which when divided by 7 gives remainder 1 are :

15,22,29, ..., 99

Here,a=15 and d=22-15=7

We have, $a_{n}=99$

nth term of an $\mathrm{AP}$ is $a_{n}=a+(n-1) d$

$\Rightarrow 99=15+(n-1) 7$

$\Rightarrow 99=15+7 n-7$

$\Rightarrow 99=8+7 n$

$\Rightarrow 7 n=99-8 $

$\Rightarrow n=\frac{91}{7}=13$

$\therefore \quad n=13$

Now, we know

$\mathrm{S}_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{13}=\frac{13}{2}[2 \times 15+(13-1) \times 7]$

$=\frac{13}{2}[30+12 \times 7]$

$=\frac{13}{2}[30+84]=\frac{13}{2} \times 114$

$=13 \times 57=741$

Question 22

A contract on construction job specifies a penalty for delay of completion beyond a certain date as follows : Rs 200 for the first day, Rs 250 for the second day, Rs 300 for the third day, etc., the penalty for each succeeding day being Rs 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work for 30 days ?

Sol :

Penalty for

First day = Rs 200

Second day = Rs 250

Third day = Rs 300

Here, a=₹ 200, d=₹ 250-200=₹ 50

n=30 days

$\therefore \mathrm{S}_{30}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{30}{2}[2 \times 200+(30-1) \times 50]$

$=15[400+29 \times 50]=15[400+1450]$

$=15 \times 1850=₹ 27750$

Question 23

Kanika was given her pocket money on 1st Jan, 2016. She puts Rs 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continued on doing so till the end of the month, from this money into her piggy bank. She also spent Rs 204 of her pocket money, and was found that at the end of the month she still has Rs 100 with her. How much money was her pocket money for the month ?

Sol :

Pocket money for Jan. 2016

Out of her pocket money, Kanika puts

Rs 1 on the first day i.e., 1 Jan.

Rs 2 on second Jan

Rs 3 on third Jan

Rs 31 on 31st Jan

Here, a=₹ 1 and d=₹ 1, n=31

$\therefore$ Amount for 31 days =1+2+3+...+31

$=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{31}{2}[2 \times 1+(31-1) \times 1]$

$=\frac{31}{2}[2+30]=\frac{31}{2} \times 32=₹ 496$

Amount spent during the period =₹ 204

Saving at the end of month =₹ 100

$\therefore$ Total savings =₹ 496+₹ 204+₹ 100=₹ 800

Question 24

Yasmeen saves Rs 32 during the first month, Rs 36 in the second month and Rs 40 in the third month. If she continues to save in this manner, in how many months will she save Rs 2000?

Sol :

Savings for the first month = Rs 32

For the second month = Rs 36

For the third month = Rs 40

Total savings for the period = Rs 2000

Here, a=₹ 32, d=36-32=₹ 4,$ \mathrm{~S}_{n}=₹ 2000$

$\mathrm{S}_{n}=₹ 2000=\frac{n}{2}[2 a+(n-1) d]$

$\Rightarrow 4000=n[2 \times 32+(n-1) \times 4]$

$\Rightarrow 4000=n[64+4 n-4]=n[60+4 n]$

$\therefore 60 n+4 n^{2}-4000=0$

$\Rightarrow n^{2}+15 n-1000=0$ (Dividing by 4)

$\Rightarrow n^{2}+40 n-25 n-1000=0$

$\Rightarrow n(n+40)-25(n-40)=0$

$\Rightarrow(n+40)(n-25)=0$

Either n+40=0, then n=-40 which is not possible being negative

or n-25=0, then n=25

$\therefore$ Required period =25 months

Question 25

The students of a school decided to beautify the school on the Annual Day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 m. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books ? What is the maximum distance she travelled carrying a flag ?

Sol :

Total number of flags = 27

To fixed after every = 2 m

The flag is stored at the middlemost flag

i.e. $\frac{27+1}{2}$ th flag i.e. at 14 th flag

$\mathrm{S}_{13}$=2[4+8+12+16+....-13 terms ]$

$=2\left[\frac{n}{2}(2 a+(n-1) d \right]$

$=2\left[\frac{13}{2}[2 \times 4+(13-1) \times 4]\right]$

$=2\left[\frac{13}{2} \times(8+48)\right] \mathrm{m}$

$=2\left[\frac{13}{2} \times 56\right]=2 \times 13 \times 28 \mathrm{~m}$

$=2 \times 364=728 \mathrm{~m}$

Maximum distance traveled in carrying a flag $=2 \times 13=26 \mathrm{~m}$