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

  Exercise 9 B

Question 1 

Ans:(i) First 15 terms of the AP : 2, 5 , 8 , 11....

Here ,  $a=2, d=5-2=3, n=15$

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

$s_{15}=\frac{15}{2}[2 \times 2+(15-1) \times 3]$

$=\frac{15}{2}[4+42]$

$=\frac{15}{2} \times 46=345$

(ii) First 50 terms of the A.P -27 , -23 , -19....

Here, a = - 27 , d =-23 -(-27)

= -23 +27

= 4 

n= 50

So  $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{50}{2}[2 \times(-27)+(50-1) \times 4]$

$S_{50}=25[-54 \times 49 \times 4]$

$=25[-54+196]$

$=25 \times 142$

$=3550$

(iii) Ans : First 17 terms of the A.P: $=\frac{1}{5}, \frac{-3}{10}, \frac{-4}{5}, \ldots$

Here, $a=\frac{1}{5}, d=\frac{-2}{10}-\frac{1}{5}$

$\frac{-3-2}{10}=\frac{-5}{10}=\frac{-1}{2}$

and $n=17$

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

$S_{17}=\frac{17}{2}\left[2 \times \frac{1}{5}+(17-1)\left(\frac{-1}{2}\right)\right]$

$=\frac{17}{2}\left[\frac{2}{5}+16\left(\frac{-1}{2}\right)\right]$

$=\frac{17}{2}\left[\frac{2}{5}-8\right]$

$=\frac{17}{2}\left[\frac{2-90}{5}\right)$

$=\frac{17}{2} \times \frac{-38}{5}$

$=\frac{-323}{5}$

$=-64 \frac{3}{2}$

(iv) Ans: First 24 terms of AP : 0 , 6, 1 , 7 , 28 ......

Here a = 0.6 d = 1.7 , 0.6 = 1.1, n = 24 

So , $S_{n}=\frac{n}{2}[2 a+(n-1) d]$

$S_{24}=\frac{24}{2}[2 \times 0.6+(24-1)(1.1)]$

$=12[1.2+23 \times 1.1]$

$=12[1.2+25.3] 84$

$=12 \times 26.5$

$=318.0$

Question 2 

Ans: (i) $\begin{aligned}=& 34+32+30+\ldots+2 \\ & Here a=34, d=32-34=-2,1=2\\ & a_{n}=a+(n-1) d \\ \Rightarrow & 2=34+(n-1) \times(-2) \\ \Rightarrow & 2-34=-2(n-1) \\ \Rightarrow & \frac{-32}{-2}=n-1 \end{aligned}$

$\Rightarrow n-1=16$

$n=16+1$

$n=17$

So , $s_{n}=\frac{n}{2}(a+l) .$

$=\frac{17}{2}(34+2)$

$=\frac{17}{2} \times 36=306$

(ii) $7+9-\frac{1}{2}+12+\ldots+67$

Here, $a=7, d=9 \frac{1}{2}-7=2 \frac{1}{2}=\frac{5}{2}, 1=67$

$U=\left(a_{n}\right)=a+(n-1) d$

$\Rightarrow 67=7+(n-1)\left(\frac{5}{2}\right)$

$67-7=\frac{5}{2}(n-1)$

$\Rightarrow \frac{60 \times 2}{5}$

$=n-1$

$\Rightarrow n-1$

$=24$

= 24n = 24 +1 

=25 

So, 

$S_{25}=\frac{n}{2}[a+1]=\frac{25}{2}[7+67]$

=$\frac{25}{2} \times 74$

$=925$

Question 3

Ans: In an AP

$d=-2, a=100,1=-10$

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

$-10=100+(n-1)(-2)$

$(n-1)(-2)=-10-100=-110$

$n-1=\frac{-110}{-2}=55$

$\Rightarrow n=55+1$

=56

$s_{56}=\frac{n}{2}[a+1]=\frac{56}{2}[100-10]$

$=28 \times 90$

$=2580$

Question 4

Ans: $\begin{aligned} \Rightarrow & \text { AP is } 54,51,48, \ldots \text { and } S_{n}=513 \\ & \text { Here, } a=54, d=51-54 \\=&-3 \end{aligned}$

$\begin{aligned} & S_{n}=\frac{n}{2}[2 a+(n-1) d] \\ & 513=\frac{n}{2}[2 \times 54+(n-1) \times(-3)] \\ \Rightarrow & 513 \times 2 \\=& n[108-3 n+3] \\ \Rightarrow & 10260 \\=& 108 n-3 n^{2}+3 n \\ \Rightarrow & 3 n^{2}-11 n+1026=0 \\ \Rightarrow & n^{2}-37 n+342=0 \\ \Rightarrow & n^{2}-18 n-19 n+342=0 \end{aligned}$

$\left\{\begin{array}{l}\because 342=-18 \times(-19) \\ -37=-18-19\end{array}\right\}$

$\begin{aligned} & \Rightarrow n(n-18)-19(n-18)=0 \\ \Rightarrow &(n-18)(n-19)=0 \end{aligned}$

Either n - 18 =0 , then n = 18

Or n- 19 =0 , then n = 19

So, Number of terms = 18 or 19 

Question 5

 

Ans: $\Rightarrow S_{9}=72, d=5$

let a be the first term,

n=9

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

$72=\frac{9}{2}[2 a+(9-1) \times 5]$

$\frac{72 \times 2}{9}=2 a+40$

$\Rightarrow 16=2 a+90$

$2 a=16-40=$

$2 a=-24$

$a=\frac{-24}{2}$

$a=-12$

 So a = -12

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

$=-12+(10-1) \times 5$

$=-12+45$

$=-33$

Question 6

Ans: In an AP, 

Sum of first 6 terms 

=42

$a_{10}: a_{30}=1: 3$

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

$S_{6}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{6}{2}[2 a+(6-1) d]$

$42=3(2 a+5 d)=6 a+15 d$

$\Rightarrow 6 a+15 d=42$

$a_{10}: a_{30}=1: 3$

$\frac{a+(10-1) d}{a+(30-1) d}=\frac{1}{3} \Rightarrow \frac{a+9 d}{a+29 d}=\frac{1}{3}$

$3 a+27 d=a+29 d$

$3 a-a=29 d-27 d$

$\Rightarrow 2 a=2 d$

$\Rightarrow a=d$

From (i) 

$6 a+15 a=42 \Rightarrow 21 a=42$

$\Rightarrow a=\frac{42}{21}=2$

$So, a=2, d=2$

So first term = 2

$a_{13}= a+(n-1) d=2+(13-1) \times 2$

$=2+12 \times 2=2+24$

$=26$

Question 7

Ans:  In a AP 

$a_{13}=4 \times a_{3}$

$a_{5}=16$

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

 So ,$\operatorname{a}_{5}=a+(n-1) d$

$=a+(5-1) d=a+4 d$

 So , $a+4 d=16$.............(i)

Similarly  

$a_{13}=a+12 d$ and $a_{3}=a+2 d$

$So, a+12 d=4 \times(a+2 d)$

$a+12 d=4 a+8 d$

$12 d-8 d=4 a-a \Rightarrow 3 a=4 d$

$a=\frac{4}{3} d$

From (i)

 $\frac{4}{3} d+4 d=16 \Rightarrow \frac{16}{3} d=16$

$\Rightarrow d=\frac{16 \times 3}{16}=3$

So, d= 3

and a = $\frac{4}{3} d=\frac{4}{3} \times 3=4$

$a=4, d=3$

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

$=\frac{10}{2}[2 \times 4+(10-1) \times 3]$

$=5(8+27)$

$=5 \times 35=175$

Question 8

Ans:  $\Rightarrow A P$ is $8,10,12, \ldots$.

Were $a=8, d=10-8=2, n=60$

So  $T_{n}=a+(n-1) d$

$\Rightarrow 60=8+(n-1) \times 2$

$\Rightarrow(n-1) \times 2=60-8$

$=52$

$n-1=\frac{52}{2}$

$=26$

$n=26+1$

$=27$

$T_{60}=a+(60-1) d$

$=8+59 \times 2$

$=8+61=69$

Sum of last 10 term $=s_{60}-s_{50}$

$=\frac{60}{2}(2 a+59 d)-\frac{50}{2}(2 a+49 d)$

$=30(2 a+59 d)-25(2 a+49 d)$

$=60 a+1770 d-50 a-1225 d$

$=10 a+545 d=10 \times 8+545 \times 2$

$=80+1090$

$=1170$

Question 9

Ans: $\Rightarrow$ In an $A P$,

$\begin{aligned}&a_{12} 0 \times T_{12}=-13 \\&s_{4}=24\end{aligned}$

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

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

$\Rightarrow-13=a+(12-1) d$

$\Rightarrow a+11 d=-13 \Rightarrow a=-13-11 d$

$s_{4}=\frac{n}{2}[2 a+(n-1) d]$

$24=\frac{4}{2}[2 a+3 d]=2[2 \times(-13-11 d)+3 d]$

$\frac{24}{2}=-26-22 d+3 d \Rightarrow 12=-26-19 d$

$\Rightarrow 12+26=-19 d \Rightarrow-19 d=38$

$d=\frac{38}{-19}=-2$

and $a=-13-11 d=-13+11 \times 2$

$=-13+22=9$

so, $a=9, d=-2$

Now , $s_{10}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{10}{2}[2 \times 9+(10-1)(-2)]$

$\begin{aligned} & 5[18+9(-2)]=5(18-18) \\=& 5 \times 0 . \\=& 0 \end{aligned}$

Question 10

Ans: $\because \Rightarrow$ Natural numbos betuteen 101 and 999 which are divisible by 2&5 both are 110 , $120,130, \ldots, 990$

Here $a,=110$ and $d=10,1=990$

 Now, $l=a_{n}=a+(n-1) d$ 

$990=110+(n-1) \times 10$

$\Rightarrow 990-110=10(n-1) \Rightarrow 10(n-1)=880$

$\Rightarrow n-1=\frac{880}{10} \Rightarrow n-1=88$

So , n = 88+ 1 = 89

Now , $s_{8 9}=\frac{n}{2}[2 a+(n-1) d]$

$=\frac{89}{2}[2 \times 110+(89-1) \times 10]$

$=\frac{89}{2}[220+88 \times 10]=\frac{89}{2}[220+880]$

$=\frac{89}{2} \times 110$

$=48950$

Question 11

Ans:  $\Rightarrow 2$ - digit number greater than 50 which are divisible by 7 , leaves a remainder of 4 are:

$53,60,67,74,81,88,95$

Here $a=53$,

$\begin{aligned} d &=7 \\ l &=95 \end{aligned}$

$\begin{aligned} \therefore a_{n}(l) &=a+(h-1) d . \\ 95 &=53+(h-1) \times 7 . \end{aligned}$

$\begin{aligned} 95-53 &=7(n-1) . \\ 42 &=7(n-1) \\ \frac{42}{7} &=(n-1) \\ 6 &=(n-1) \\(n-1) &=6 \\ n &=6+1 \\ n &=7 . \end{aligned}$

Then , 

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

$=\frac{7}{2}[2 \times 53+(7-1) \times 7]$

$=\frac{7}{2}[106+6 \times 7]$

$=\frac{7}{2}[106+42]$

$=\frac{7}{2} \times 48$

$=518$

Question 12

Ans: (i) Integers between 100 and 200 which are divisible by 9 are 108, 117 , 126 ,......198

Here 

a= 108

d =9

and l = 198

$\begin{aligned} \therefore a_{n}(l) &=a+(n-1) d . \\ 198 &=108+(n-1) \times 9 . \\ 198-108 &=9(n-1) \\ 90 &=9(n-1) \\ \frac{90}{9} &=n-1 \\ 10 &=n-1 \\ 10+1 &=n \\ n &=11 \\  & \end{aligned}$

Then 

$\begin{aligned} S_{11} &=\frac{h}{2}[2 a+(h-1) d] . \\ &=\frac{11}{2}[2 \times 108+(11-1) \times 9] \\ &=\frac{11}{2}[216+10 \times 9] \\ &=\frac{11}{2}[216+90] \end{aligned}$

$\frac{11}{2}\times 306$

=1683

(ii) Then sum of integers from 101 to 199

Here, a = 101,

d= 1 

l = 199

and n = 99

$s_{n}=$ $\frac{n}{2}[2 a+(n-1)d]$

$=\frac{99}{2}[2 \times 101+(99-1) \times 1]$

$=\frac{99}{2}[202+98]$

$=\frac{99}{2}\times 300$

=14850

∴ Sum of integers which are not divisible by 9 

$=14850-1683$

$=13167$

Question 13

Ans:  Number between 9 and 95 when divided by 3, 

leaves remainder 1 

$10,13,16,19, \ldots . .99$

Here, 

$\begin{aligned} q &=10, \\ d &=3, \\ \text { and } l &=94 . \end{aligned}$

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

$94=10+(n-1) \times 3$

$94-10=3(n-1) .$

$84=3(n-1)$

$\frac{84}{3}=n-1$

$28=n-1$

$28+1=n$

$29=$n

$n=29$

Then, middle term 

$\begin{aligned} &=\frac{29+1}{2} \text { th } \\ &=\frac{30}{2} \mathrm{th} . \end{aligned}$

=15th term

Sum of first 14 terms 

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

$=\frac{74}{2}[2 \times 10+(14-1) \times 3]$

$=7[20+13 \times 3]$

$=7[20+39]$

$=7 \times 59$

$=413$

Middle terms = 10+15\times 3

= 10+45 = 55

After middle term; number of terms = 14 

Whose first terms (a)= 55 , d = 3

and n = 14

$\therefore s_{14}=\frac{19}{2}[2 \times 55+(14-1) \times 3]$

$\begin{aligned} &=7[110+39] \\ &=7 \times 149 \\ &=1043 \end{aligned}$

Question 14

Ans: $S_{n}=$ Sum of first n terms of an A.P.

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

$To$ prove $=S_{12}=3\left(S_{8}-S_{4}\right)$

from $R \cdot H \cdot S=3\left[S_{8}-S_{4}\right]$

$=3\left[\frac{8}{2}[2 a+(8-1) d]-\frac{4}{2}[2 a+(4-1) d]\right] .$

$=3\left[\frac{8}{2}(2 a+7 d)-\frac{4}{2}[2 a+3 d)\right]$

$=3[4(2 a+7 d)-2(24+3 d)]$

$=3[84+28 d-4 a-6 d]$

$=3[4 a+22 d]$

$=12 a+66 d .$

$=6[2 a+11 d) .$

$\frac{12}{2}[2 a+(12-1) d]$= $S_{12}$ =L.H.S

Hence proved 

Question 15

Ans: Sum of first n even natural number $=\left(1+\frac{1}{h}\right)$ 

(Sum of first n odd natural numbers)

Sum of first n even natural number (2,4,6,8......2n)

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

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

$=\frac{n}{2}[4+2n-2] .$

$=\frac{n}{2}[2+2n] .$

$=\frac{n}{2} \times 2(1+n)$

$=n(1+n)$

and Sum of n odd natural number (1, 3,5 ,.......(2n -1))

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

$=\frac{n}{2}[2+2 n-2]$

$=\frac{n}{2}[2 n]$

$=\frac{n}{2} \times 2[n]$

$=n^{2} .$

Then , 

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

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

$=n^{2} \times \frac{n+1}{n}$

$=n(n+1)$

Question 16

Ans: Given,

First term $\left(a_{1}\right)=5$

Common difference $\left(d_{1}\right)=36$.

$S_{n}=$ Sum of $2_{n}$ terms of another $A P$ whose first term $\left(a_{2}\right)=36$ and common difference $\left(d_{2}\right)=5$.

 In first $A P$

$\begin{aligned} S_{n} &=\frac{n}{2}[2 a+(n-1) d] . \\ &=\frac{n}{2}[2 \times 5+(n-1) \times 36] \\ &=\frac{n}{2}[10+36 n-36] \\ &=\frac{n}{2}[36 n-26] . \end{aligned}$

$=\frac{n}{2} \times 2[18 n-13]$

$=n[18 n-13]$

In Second AP, 

$S_{2 N}=\frac{2 n}{2}[2 \times 36+(2 n-1) \times 5]$

$=n[72+10{n}-5]$

$=\mathrm{n}[67+10 \mathrm{n}]$

$\therefore S_{n}=S_{2 n}$

18 (18n- 13) = n(67 + 10n)

18 n - 13 =  $\frac{n}{n}(67+10 n)$

$18 n-10 n=67+13$

8n = 80

n = $\frac{80}{8}$

n=10

Question 17

Ans: Sum of first 10 terms of an AP = $4 \times sum$ of first 5 terms 

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

$\begin{aligned} \therefore S_{n} &=\frac{n}{2}[2 a+(n-1) d] \\ S_{10} &=\frac{10}{2}[2 a+(10-1) d] \\ &=5[2 a+9 d] \\ &=10a+45d \end{aligned}$

$\begin{aligned} S_{{5}} &=\frac{5}{2}[2 a+(5-1) d] \\ &=\frac{5}{2}[2 a+4 d] \end{aligned}$

$=\frac{5}{2} \times 2[a+2 d]$

$=5 a+10 d$

According to the question,

10a + 45 d =  $4 \times(5 a+10 d)$

$10 a+45 d=20 a+40 d .$

$45 d-40 b=20 a-10 a$

$5 a=109 .$

$\frac{5}{10}=\frac{a}{d}$

$\frac{a}{d}=\frac{5}{10}$

$\frac{a}{d}=\frac{1}{2}$

∴ Ratio in first term to the common difference = 1:2

Question 18

Ans:  Given, 

In the first row, plants are = 37

In second row = 35 

In third row = 33

and in the last row = 5

Here , 

a= 37 

d= 35 - 37 

d= - 2

$\begin{aligned} a_{n}(1)=& a+(n-1) d . \\ 5=& 37+(n-1) \times(-2) \\ 5 &-37=-2 n+2  \\ &-32=-2(n-1) . \\ & \frac{+32}{+2}=n-1 \end{aligned}$

$16=n-1$

$16+1=n$

$17=n$

$n=17$

So, number of rows = 17 

Then, 

Total plants $\left(S_{n}\right)=\frac{n}{2}[24+(n-1) d]$

$=\frac{17}{2}[2 \times 37+(17-1) \times(-2)] .$

$=\frac{17}{2}[74-32]$

=357

Hence , the total plants is 357

Question 19

Ans: Given , 

Total logs = 200 

In the bottom row, number of logs = 20 

and next row above it = 19 

Next row above it = 18

And Soon 

- AP is 20,19,18,17,16....... and $S_{n}=200$

Let the top row be the nth row 

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

and $S_{n}=200$

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

$200=\frac{n}{2}[2 \times 20+(n-1) \times(-1)]$

$400=n[40-n+1]$

$400=40 n-n^{2}+h$

$400=41 n-n^{2}$

$n^{2}-41 n+400=0$

$n^{2}-25 n-16 n+400=0$

$n(n-25)-16(n-25)=0$

$(n-25)(n-16)=0$

n -25 =0 or n-16=0

n=25 or n =16

But n = 25 is not possible 

 As number of logs in the first row = 20 

and d = - 1 

Number of row = 16 

and number of log in 16th row 

$=9+(n-1) d$

$=20+(16-1) \times(-1)$

$=20-15$

$=5 logs$

Question 20

Ans: Giver,

Total amount= ₹1590,

Number of cash prizes = 7 

and each prize is Rs 50 less than the proceeding prize 

Let first prize = Rs a 

Then second prize = a - 50 

Third prize = a - 100

And so on 

$\therefore$ first term $=a$,

common difference (d) $=-50$,

$S_{n}=1890$

$n=7$

$\begin{aligned} S_{n}=& \frac{n}{2}[2 a+(n-1) d] \\ 1890=& \frac{7}{2}[2 a+(7-1) \times(-50)] \\ 3780=& 7[2 a-300] \\ & \frac{3780}{7}=2 a-300 \\ 540=& 2 a-300 \\ 540+300=& 2 a \\ 840=& 2 a \\ \frac{840}{2}=9 \\ 420=a \\ a=420 \end{aligned}$

Hence prizes are $₹420, ₹ 370, ₹ 320 ; 270, ₹ 220, ₹ 170$ and $₹ 120$.