S Chand CLASS 10 Chapter 9 Arithmetic and Geometric Progression Exercise 9 A

 Exercise 9 A

Question 1 

Ans: (i) Yes 

(ii) Yes, it is on arithmetic progression as cost of digging in the beginning is Rs 125 for the first meter and then increasing Rs 65 for every subsequent meter. 

Question 2

Ans: (i) a= 10 , d= 7 

10 , 17 , 24 ,31 , 38 

(ii) a = 55 , d = -4

So AP =55, 51 47, 43

(iii) a =-1 , d=  $\frac{1}{2}$

$A P=-1, \frac{-1}{2}, 0, \frac{1}{2}, 1$

(iv) $a=-3: 25, d=-0.25$

$A P=-3.25,-3.50,-3.75,-4.90,-4.25$

Question 3

Ans:  $\Rightarrow$ List of numbers : $-12,-9,-6,-3,0,3$

 Here $a=-12, d=-9-(-12)=-9+12=3$ 

$a=-12, d=3$

(ii)  $\Rightarrow$ In an AP, a $=3, d=0, n=7$,

$\therefore a_{n}=a+(n-1) d=3+(7-1)+0$

$=3+0=3$

Question 4

Ans: (i) $\Rightarrow A P=-5,-\frac{5}{2}, 0, \frac{5}{2}, \ldots$

$\therefore, a=-5, d=\frac{-5}{2}-(-5)$

$=\frac{-5}{2}+5=\frac{5}{2}$

$So , a_{25}=a+(n-1) d=-5+(25-1) \times \frac{5}{2}$

$=-5+24 \times \frac{5}{2}$

$=-5+60$

$=55$

(ii) AP is 27, 22, 17 ....

∴ a = 27 , d = (22- 27 ) = -5

So $a_{30}=a+(n-1) d$

$=27+(30-1) \times(-5) .$

$=27+29(-5)$

$=27-145=-118$

(ii) $\Rightarrow A P=-0.1,-0.2,-0.3, \ldots$

$\therefore a=-0.1$

$d=-0.2-(-0.1)=-0.2+0.1=-0.1$

$So , a_{10}=a+(n-1) d=-0.1+(10-1)(-0.1)$

$=-0.1+9 \times(-0.1)=-0.1-0.9=-1$

Question 5

Ans:  K , 2 k - 1 & 2K + 1 are the three Terms of an AP, then 

$a=k, d=2 k-1-k=k-1$ (i)

and $d=2 k+1-(2 k-1)=2 k+1-2 k+1$

$=2$ (ii)

From (i) and (ii)
So , K-1 = 2 = k = 2+1 = 3

SO k = 3 

Question 6

Ans: $\Rightarrow A p$ is $\frac{1}{3 q}, \frac{1-6 q}{3 q}, \frac{1-12 q}{3 q}$

$d=\frac{1-6 q}{3 q}-\frac{1}{3 q}=\frac{1-6 q-1}{3 q}=\frac{-6 q}{3 q}$

$=-2$

So , Common difference = -2

Question 7

Ans: = Which term of Ap 5, 13 , 21 , ...... is 181 

Let it be nth term, then 

In AP a = 5, d = 13 -5 = 8

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

$181=5+(n-1) \times 8$

$\Rightarrow|8|=5+8 n-8$

$\Rightarrow 8 n=181-5+8=184$

$\Rightarrow n=\frac{184}{8}=23$

so, 181 is 23 rd term.

Question 8

Ans:

 $\begin{aligned}&\Rightarrow \text { Ap is } 3,10,17, \ldots \\&\therefore a=3, d=10-3=7\end{aligned}$

Let nth term is 84 more than its 13 the term

$\begin{aligned}&a_{13}=a+(n-1) d=3+(13-1) \times 3 \\&=3+12 \times 7=87 \\&a_{n}=3+(n-1) \times 7=3+7 n-7=7 n-4 \\&\text { Now, } 7 n-4=84+87=171 \\&7 n=171+4=175 \\&n=\frac{175}{7} \\&=25\end{aligned}$

So, 25 th is the required term.

Question 9

Ans: In an AP 

6th term = 19 and 

16th term = 15 + 11th term 

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

So , $a_{6}=a+(n-1) d$

$\Rightarrow 19=a+5 d$.......(i)

$\Rightarrow a+5 d=19$

similarly

$a_{11}=a+10 d$ and $a_{16}=a+15 d$

Now, $a+15 d=a+10 d+15$

$15 d-10 d=15$

$\Rightarrow 5 d=15$

$\Rightarrow d=\frac{15}{5}=3$

Erom (i)

$a+5 \times 3=19$

$\Rightarrow a+15=19$

$\Rightarrow a=19-15=4$

$or=4, d=3$

So, $A P$ is $4,7,10,13,16 \ldots$

Question 10

Ans:   $\Rightarrow$ 2nd term $+7$ th term $=30$

15 th term $=2 \times 8$ th term $-1$

let a be the first term and $d$ be the common difference, then

$a_{2}=a+(n-1) d=a+(2-1) d=a+d$

similarly, $a=a+(7-1)$

$d \Rightarrow a+6 d$

$a_{15}=a+(15-1)$

$d \Rightarrow a+14 d$ and

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

$\Rightarrow a+7 d$

Now, $a_{2}+a_{7}=30$

$\Rightarrow a+d+a+6 d=30$

$\Rightarrow 2 a+7 d=30$

and $a+14 d=2 \times(a+7 d)-1$

$a+14 d=2 a+14 d-1$

$2 a-a=1$

$\Rightarrow a=1$

From (i), $2 \times 1+7 d=30$

$\Rightarrow 2+7 d=30$

$\Rightarrow 7 d=30-2=28$

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

So , d = 4 a = 1

So , AP  = 1, 5 ,9 , 13 , 17.....

Question 11

Ans:  In an Ap 

4th term $(a 4)$

=11

5 th term $+7$ th iterm

$=34$

let a be the first term and $d$ be the common difference, then

$\begin{aligned} & a_{4} \\=& a+(n-1) d \\=& a+(4-1) d \\ \Rightarrow & a+3 d=11 \end{aligned}$

similarly.

$a_{7}=a+(7-1) d \Rightarrow a+6 d$

We know that, $a_{4}+a_{7}=34$

$\therefore a+4 d+a+6 d=34$

$\Rightarrow 2 a+10 d=34$

$a+5 d=17$

subtracting (ii) from (i).

$\begin{aligned}&=(a+3 d)-(a+5 d)=11-17 \\&=a+3 d-a-5 d=-6 \\&-2 d=-6\end{aligned}$

$2 d=6 \Rightarrow d=\frac{6}{2}=3$

$\therefore$ Common difference $=3$

Question 12

Ans: $\Rightarrow A P=213,205,197, \ldots .37$

let 37 be the $n$th term

Now, $a=213 \mathrm{~A} d=-205-213=-8$

$\therefore a_{n}=a+|n \cdot 1| d$

$\Rightarrow 37=213+(n-1)(-8)$

$37=213-8 n+8$

$n=\frac{184}{8}=23$

There are 23 term in the AP 

Middle term = $\frac{23+1}{2}=12$ th term

$a_{12}=a+11 d$

$2213+11(-8)=213-88=125$

Question 13 

Ans: In an AP 

3rd term = 4

9th term =- 8 

Which term is 0 

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

$\begin{array}{rl} & a_{3}=a+(n-1) d \\ \Rightarrow & 4=a+(3-1) d \\ \Rightarrow & a+2 d=4 \\ \text { similarly, } a+8 d=-8 \\ \text { subtracting (i) from (ii), } \\ \Rightarrow(a+8 d)-(a+2 d)=-8-4 \\ 6 & d=-12 \Rightarrow d=\frac{-12}{6}\end{array}$

=-2

and $a+2(-2)=4$

$\Rightarrow a-4=4$

$\Rightarrow a=4+4=8$

$\therefore a=8, d=-2$

Let $a_{n}$ be equal to 0 , then

$\begin{aligned} & a+(n-1) d=0 \\ \Rightarrow & 8+(n-1)(-2)=0 \\ \Rightarrow &+8-2 n+2=0 \Rightarrow 10-2 n=0 \end{aligned}$

$\Rightarrow 2 n= 10$

$n=\frac{10}{x}$

$=5$

=5th  term is zero

Question 14

Ans: $\Rightarrow$ In an AP

8 th $\operatorname{tern}\left(a_{8}\right)=0$

To prove that 38 th term

$=3 \times 18$ th term

$=$ Let $a$ be the first teron and $d$ be the common

$=$ difference, then

$\begin{aligned}&a_{8}=a+(n-1) d \\&\Rightarrow a+(8-1) d=0 \\&\Rightarrow a+7 d=0 \\&\Rightarrow a=-7 d\end{aligned}$

$\begin{aligned}&\text { Similarly } \\&a_{18}=a+17 d \\&a_{38}=a+37 d\end{aligned}$

Now , a + Ad= -7d + 17d 

 = 10 d  and a + 37 d 

= - 7d + 37 d 

= 30 d 

So , 30 d 

$=3 \times 10 \mathrm{~d}$

$=30 \mathrm{~d}$

$= a_{38}=3 \times a_{18}$

Question 15

Ans: AP is 120 , 116 , 112.....

Which first term of this Ap will be negative 

Here , a =120, d = 116 - 120 =-4 

Let nth term of the given AP be the negative term  $a_{n}$ or $T_{n}<0$

$\Rightarrow a+(n-1) d<0$

$\Rightarrow 120+(n-y)(-4)<0$

$\Rightarrow 120-4 n+4<0$

$\Rightarrow 124-4 n<0 .$

$\Rightarrow 4 n>124$

$\Rightarrow n>\frac{124}{4}$

= 31

So', $n=32$ be the first negative term

$\Rightarrow 32$ nd term is the first negative term

Question 16

Ans: $\Rightarrow 3$-digits terms are $100,101,102, \ldots, 999$ and terms which are divisible by 7 Will be $105,112,119, \ldots, 994$

$\because a=105, d=7$

Last term ($a_{n}$) = 994

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

$994=105+(n-1) \times 7$

$=105+7 n-7=7 n+98$

$7 n=994-98$

$=896$

$n=\frac{896}{7}=128$

Question 17

Ans:- Let $d$ be the common difference of the two $A P$ series $\alpha_{1}$ and $a_{2}$ are the first term of the two AP's respectively.

In first $A P, a_{1}=1, a_{2}=-8$

Difference between their 4 th terms

$=\left(a_{1}+3 d\right)-\left(a_{2}+3 d\right)-$

$=-1-(-8)$

$a_{1}+3 d-a_{2}-3 d$

$=-1+8$

$\Rightarrow a_{1}-a_{2}=7$

$or a_{2}+3 d-a_{1}-3 d$

$=-8-(-1)$

$a_{2}-a_{1}=-8+1=-7$

$\therefore$ Difference $=7$ or $-7$

Question 18

Ans:  $\Rightarrow$ In an AP

Ratio between 4 th term and 9 th term $=1: 3$

let a be the first term and $d$ be the common difference, then

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

$=a+(n-1) d$

$=a+(4-1) d$

$=a+3 d$

Similarly 9th term = a + 8 d 

Now,

So, $\frac{a+3 d}{a+8 d}=\frac{1}{3}$

$\Rightarrow 3 a+9 d=a+8 d$

$\Rightarrow 3 a-a$

$=3 d-9 d$

$\Rightarrow 2 a=-d$

$\Rightarrow d=-2 a$

Now, $a_{12}=a+(12-1) d$

$=a+11 d$

and $a_{5}=a+(5-1) d$

$=a+4 d$

So,$\frac{a+11 d}{a+4 d}$

$\frac{a+(-2 a) \times 11}{a+4(-2 a)}$

$=\frac{a-22 a}{a}-8 a$

$=\frac{-21 a}{-7 a}$

$=\frac{3}{1}$

$=3: 1$

Question 19

Ans: 7th term from the end of AP 

7, 10 , 13, ......184 

Hence , a = 7, d =10-7 = 3

$\begin{aligned} & T_{n}(1)=184 \\ T_{n}=a+(n-1) d \\ \Rightarrow & 7+(n-1) \times 3 \\ \Rightarrow & 7+3 n-3 \\=& 4+3 n \end{aligned}$

$\operatorname{So} 4+3 n=184$

$\Rightarrow 3 n=184-4$

$=180$

$n=\frac{180}{3}=60$

7th term from the last will be 60- (7-1)

60 - 6 = 54th from the beginning 

So $T_{54}=a+53 d$ 

$\begin{array}{rl} & =7+53 \times 3 \\  & 7+159 \\ = & 166\end{array}$

Question 20

Ans: Four angles of a quadrilateral are in AP 

Sum of the first 3 angles = 2 $\times$ 4th angle

Adding 4th angle to both sides 

Sum of 4 angle $=2 \times 4th angle +4th angle

$=3 \times 4$th angles

But Sum of 4 angles of a quadrilateral = 360 

So,  $=3 \times 4$th angles

4 th angle $=\frac{360^{\circ}}{3}=120^{\circ}$

$\Rightarrow 4$ th angle $=120^{\circ}$

Let a be the first term (angle) and d be the common difference 

So, a + 3d = 120 .............(i)

And $a+a+d+a+2 d=360^{\circ}-120^{\circ}=240$

$3 a+3 d=240$..............(ii)

Subtracting 2a= 120 = $a=\frac{120^{\circ}}{2}=60^{\circ}$

But $a+3 d=120$

$\Rightarrow 60+3 d= 120 \Rightarrow 3 d$

$=120-60=60$

$d=\frac{60^{\circ}}{3}=20^{\circ}$

So, Angles are  $60^{\circ}, 80^{\circ}, 100^{\circ}, 120^{-}$

Question 21

Ans: Sum of 3 numbers in $A P=-3$ 

and product $=8$ 

let three numbers in AP be 

$a-d, a, a+d$

$a-d+a+a+d$

=-3

$\Rightarrow 3 a=-3$

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

$=-1$

and (a- d) $a(a+d)=8$

$\left(a^{2}-d^{2}\right) a=8$

$\left.\left[(-1)^{2}-d^{2}\right)\right](-1)=8$

$1-d^{2}=-8$

$d^{2}=1+8$

$d_{2}=9$

$d^{2}=(+3)^{2}$

So, Number be -1 , -3 ,-1 -1, +3

= -4 , -1 , 2 

Or -1+3, -1, -1 , -3

= 2 , -1 , -4 

Question 22

Ans: Ram prasad's savings in first week = 10 rs 

Then his weekly saving is increasing = Rs 2.75 

So, a = Rs 10 and d = Rs 2.75 

nth week saving = rs 59.50

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

$59.50=10+(n-1)(2.75)$

$59.50-10.00=(n-1)(2.75)$

$49.50=(n-1)(2.75)$

So, $n-1=\frac{49.50}{2.75}$

$=18$

so, $n=18+1$

$=19$

Question 23

 

Ans: $\Rightarrow$ Eer an AP

$T_{p}+T_{p}+2 q=2 T_{p+q}$

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

$T_{p}=a+(p-1) d$,

$T_{p}+2 q=a+(p+2 q-1) d$

$T_{p}+q=a+(p+q-1) d$

Now, LHS $=T_{p}+T_{p+2 q}$

$=a+(p-1) d+a+(p+2 q-1) d$

$=a+d p-d+a+p d+2 q d-d$

$=2 a+2 d p+2 q d-2 d$

$=2(a+d p+q d-d)$

$=2[a+(p+q-1) d]$

$=2 \times T_{p+q}$

=R H S