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 $₹ 120$.
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