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-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
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