NCERT Exemplar Class 10 Maths Solutions Chapter 5 - Arithmetic Progressions

Question:1

In an AP, if d = –4, n = 7, a_n=4, then a is

(A) 6

(B) 7

(C) 20

(D) 28

Answer:

Given
$a_{\text{n}} = 4, d = - 4, n = 7$

$Use \ a_{\text{n}}= a + (n - 1).d$

$a + (7 - 1). (- 4) = 4$

$a +(6).(-4) = 4$

$a - 24 = 4$

$a = 4 + 24$

$a = 28$
Hence, option D is correct

Question:2

In an AP, if a = 3.5, d = 0, n = 101, then a_n will be
(A) 0
(B) 3.5
(C) 103.5
(D) 104.5

Answer:

We know that
$ a_{\text{n}} = a + (n -1).d$

$Given, a = 3.5, d = 0, n = 101$

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

$a_{\text{n}} = 3.5 + (101 - 1).0$

$a_{\text{n}} = 3.5 + 0$

$a_{\text{n}} = 3.5$
Hence, option B is correct.

Question:3

The list of numbers – 10, – 6, – 2, 2... is
(A) an AP with d = – 16
(B) an AP with d = 4
(C) an AP with d = – 4
(D) not an AP

Answer:

$a_1 = -10, \quad a_2 = -6, \quad a_3 = -2, \quad a_4 = 2$

$a_2 - a_1 = -6 - (-10)$

$= -6 + 10$

$= 4$

$a_3 - a_2 = -2 - (-6)$

$= -2 + 6$

$= 4$

$a_4 - a_3 = 2 - (-2)$

$= 2 + 2$

$= 4$

Since the given sequence is an AP with d = 4
ANS - (B)

Question:4

The 11th term of the AP:
$-5, \frac{-5}{2},0,\frac{5}{2}, $ \ldots $ ..is\\$
(A) –20
(B) 20
(C) –30
(D) 30

Answer:

Here AP is

$
-5,-\frac{5}{2}, 0, \frac{5}{2}, \ldots \ldots \ldots
$
Here

$
a_1=-5, a_2=-\frac{5}{2}
$

(Common difference) $d=a_2-a_1$

$
\begin{aligned}
& =-\frac{5}{2}-(-5) \\
& =-\frac{5}{2}+5 \\
& =\frac{-5+10}{2} \\
& =\frac{5}{2}
\end{aligned}
$
$
\begin{aligned}
& a_{\mathrm{n}}=a_1+(n-1) \cdot d \\
& \text { Put } n=11 \\
& a_{11}=a_1+(11-1) \cdot d \\
& a_{11}=-5+(10) \cdot(5 / 2) \\
& =-5+5 \cdot(5) \\
& =-5+25 \\
& =20
\end{aligned}
$

Hence, option B is correct.

Question:5

The first four terms of an AP, whose first term is –2 and the common difference is –2, are
(A) – 2, 0, 2, 4
(B) – 2, 4, –8, 16
(C) – 2, – 4, – 6, – 8
(D) – 2, – 4, – 8, –16

Answer:

Given,
$a = - 2, d = - 2$

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

$a + 2d = - 2 + 2(-2) =-2 - 4 = - 6$

$a + 3d = - 2 + 3(-2)\\ =-2 - 6 \\ = - 8\\$
The given AP is -2, -4, -6, 8
Hence, option (C) is correct.

Question:6

The 21st term of the AP, whose first two terms are 3 and 4, is
(A) 17
(B) 137
(C) 143
(D) –143

Answer:

Given

$a_1 = -3$
$a_2 = 4$

(Common difference) $d = a_2 - a_1$
$= 4 - (-3)$
$= 4 + 3$
$= 7$

We know that the general term of an arithmetic progression is:
$a_n = a + (n - 1)d$

Put $n = 21$:
$a_{21} = a + (21 - 1)d$
$= -3 + 20 \cdot 7$
$= -3 + 140$
$= 137$

$\boxed{a_{21} = 137}$

Hence, option B is correct.

Question:7

If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
(A) 30
(B) 33
(C) 37
(D) 38

Answer:

We know that

$a_n = a + (n - 1)d \tag{1}$

Given: $a_2 = 13$, $a_5 = 25$

From equation (1), put $n = 2$:
$a + (2 - 1)d = 13$
$\Rightarrow a + d = 13 \tag{2}$

Put $n = 5$ in equation (1):
$a + (5 - 1)d = 25$
$\Rightarrow a + 4d = 25 \tag{3}$

Subtract equation (2) from equation (3):
$(a + 4d) - (a + d) = 25 - 13$
$\Rightarrow 3d = 12$
$\Rightarrow d = \frac{12}{3} = 4$

Substitute $d = 4$ in equation (2):
$a + 4 = 13$
$\Rightarrow a = 13 - 4 = 9$

Now find $a_7$:
$a_7 = a + (7 - 1)d$
$= 9 + 6 \cdot 4$
$= 9 + 24 = 33$

$\boxed{a_7 = 33}$

Hence, option B is correct.

Question:8

Which term of the AP: 21, 42, 63, 84... is 210?
(A) 9th
(B) 10th
(C) 11th
(D) 12th

Answer:

$a_n=210 \mathrm{AP}$ is $21,42,63,84, \ldots$

Here $a_1=21 d=a_2-a_1 \quad($ where $d$ is the common difference $)$

$d=42-21 \quad\left(a_1\right.$ is the first term $)$

$d=21$

$a_n=210 $

$a_1+(n-1) \cdot d=210$

$21+(n-1) \cdot 21=210$

$21+21 n-21=210$

$21 n=210$

$n=\frac{210}{21}$

$n=10$

Hence the $10^{\text {th }}$ term of the AP is 210

Question:9

If the common difference of an AP is 5, then what is $a_{18} - a_{13}$?
(A) 5
(B) 20
(C) 25
(D) 30

Answer:

Given

$
\begin{aligned}
& d=5 \\
& a_{18}-a_{13}=?
\end{aligned}
$

We know that

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

Put $n=18$,

$
a_{18}=a+(18-1) \cdot d
$

Put $n=13$,

$
a_{13}=a+(13-1) \cdot d
$

Now,

$
\begin{gathered}
a_{18}-a_{13}=(a+(18-1) \cdot d)-(a+(13-1) \cdot d) \\
=a+17 d-a-12 d \\
=17 d-12 d \\
=5 d \\
=5 \times 5 \\
=25
\end{gathered}
$
Option (C) is correct.

Question:10

What is the common difference of an AP in which $a18 - a14 = 32$?
(A) 8
(B) – 8
(C) – 4
(D) 4

Answer:

Given,

$
a_{18}-a_{14}=32
$
We know that $\mathrm{a}_{\mathrm{n}}=a+(n-1) \cdot d$
Put $\mathrm{n}=18$

$
\begin{aligned}
& a_{18}=a+(18-1) \cdot d \\
& a_{18}=9+17 d \ldots \ldots \ldots \ldots \ldots \ldots(1) \\
& n=14 \\
& a_{14}=a+(14-1) \cdot d \\
& a_{14}=9+13 d \ldots \ldots \ldots \ldots \ldots(2) \\
& \text { subtract eq }(2) \text { from eq(1) } \\
& a_{18}-a_{14}=32 \\
& (a+17 d)-(a+13 d)=32 \\
& a+17 d-a-13 d=32 \\
& 4 d=32 \\
& d=\frac{32}{4} \\
& d=8
\end{aligned}
$
Hence common difference is 8.

Question:11

Two APs have the same common difference. The first term of one of these is 1, and that of the other is 8. Then the difference between their 4th terms is
(A) –1
(B) – 8
(C) 7
(D) –9

Answer:

Let the first AP be $a_1, a_2, a_3, \ldots \ldots \ldots$.
Let the second AP be $b_1, b_2, b_3, \ldots \ldots \ldots \ldots$
Given, $a_1=-1, \quad b_1=-8$
Difference between their $4_{\text {th }}$ terms is:

$
a_4-b_4=\left(a_1+3 d\right)-\left(b_1+3 d\right)
$

Given that the common difference of both A.Ps is equal,

$
\begin{gathered}
a_4-b_4=-1-(-8) \\
=-1+8 \\
=7
\end{gathered}
$

Hence, option (C) is correct

Question:12

If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be
(A) 7
(B) 11
(C) 18
(D) 0

Answer:

Given $7 \times\left(a_7\right)=11 \times\left(a_{11}\right)$

$
\begin{gathered}
7 \times(a+6 d)=11 \times(a+10 d) \quad\left(\text { Using } a_{\mathrm{n}}=a+(n-1) \times d\right) \\
7 a+42 d=11 a+110 d \\
11 a-7 a+110 d-42 d=0 \\
4 a+68 d=0
\end{gathered}
$

Dividing by 4 , we get

$
\begin{gathered}
a+17 d=0 \quad \ldots \\
a_{18}=a+(18-1) d \\
=a+17 d \\
=0
\end{gathered}
$

Hence, option D is correct.

Question:13

The 4th term from the end of the AP: –11, –8, –5, ..., 49 is
(A) 37
(B) 40
(C) 43
(D) 58

Answer:

Given

$
\begin{gathered}
7 \times\left(a_7\right)=11 \times\left(a_{11}\right) \\
7 \times(a+6 d)=11 \times(a+10 d) \quad\left(\mathrm{Using} a_{\mathrm{n}}=a+(n-1) \times d\right) \\
7 a+42 d=11 a+110 d \\
11 a-7 a+110 d-42 d=0 \\
4 a+68 d=0
\end{gathered}
$

Dividing by 4 , we get

$
\begin{gathered}
a+17 d=0 \quad \ldots \\
a_{18}=a+(18-1) d \\
=a+17 d \\
=0
\end{gathered}
$

Hence, option B is correct.

Question:14

The famous mathematician associated with finding the sum of the first 100 natural numbers is
(A) Pythagoras
(B) Newton
(C) Gauss
(D) Euclid

Answer:

[C]
Famous mathematician Gauss found the sum of the first 100 natural numbers.
According to this when he added the first and the last number in the series, the second and the second last number in the series, and so on. He gets a sum of 101 and50 pairs exist so the sum of 100 natural numbers is 50×101=5050.

Question:15

If the first term of an AP is i–5 and the common difference is 2, then the sum of the first 6 terms is
(A) 0
(B) 5
(C) 6
(D) 15

Answer:

Given $a=-5, d=2$ (where $a$ is the first term and $d$ is the common difference)

$
\begin{gathered}
S_6=\frac{6}{2}(2 a+(6-1) d) \\
=3(2 \times(-5)+5 \times 2) \\
=3(-10+10) \\
=3(0)=0
\end{gathered}
$
Hence, option A is correct

Question:16

The sum of the first 16 terms of the AP: 10, 6, 2... is
(A) –320
(B) 320
(C) –352
(D) –400

Answer:

Given: $\mathrm{AP}=10,6,2 \ldots$ here

$
\begin{aligned}
& a_1=10, a_2=6 \\
& d=a_2-a_1 \\
& d=6-10 \\
& d=-4 \\
& S_{16}=\frac{16}{2}(2 a+(16-1) d) \\
& =8(2 \times 10+15 \times(-4)) \\
& =8(20-60) \\
& =8 \times(-40)=-320
\end{aligned}
$
Hence, option A is the correct answer.

Question:17

$\text{In an AP if a = 1,}$ $a_{\text{n}} = 20 $ and $ S_{n} = 399$\text{, then n is }$
(A) 19
(B) 21
(C) 38
(D) 42

Answer:

Given : $\mathrm{a}=1, \mathrm{a}_{\mathrm{n}}=20, \mathrm{~S}_{\mathrm{n}}=399$

$
\begin{aligned}
& S_n=\frac{n}{2}\left(a+a_n\right) \\
& 38=n
\end{aligned}
$
He, once option C is correct.

Question:18

The sum of the first five multiples of 3 is
(A) 45
(B) 55
(C) 65
(D) 75

Answer:

Given:- Multiples of 3 are: $3,6,9,12, \ldots \ldots$.

$
\begin{gathered}
a_1=3, \quad d=a_2-a_1 \\
d=6-3 \\
d=3 \\
S_5=\frac{5}{2}\left[2 a_1+(5-1) d\right] \\
=\frac{5}{2}(2 \times 3+4 \times 3) \\
=\frac{5}{2}(6+12) \\
=\frac{5}{2} \times 18
\end{gathered}
$
Option A is correct.

Question:1

i)Which of the following forms an AP? Justify your answer.
–1, –1, –1, –1, ...
ii)
Which of the following forms an AP? Justify your answer.
0, 2, 0, 2, ...
iii)
Which of the following is an AP? Justify your answer.
1, 1, 2, 2, 3, 3,...
iv)
Which of the following forms an AP? Justify your answer.
11, 22, 33,...
v)
Which of the forms form an AP? Justify your answer.
$\frac{1}{2}, \frac{1}{3}, \frac{1}{4},......................$
vi)
Which of the following forms an AP? Justify your answer.
$2, 2^2, 2^3, 2^4, ...$
vii)
Which of the following forms an AP? Justify your answer.
$\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots$

Answer:

i)
Answer. [True]
Solution.

$-1,-1,-1,-1, \ldots$ here $a_1=-1, a_2=-1, a_3=-1, a_4=-1$

$
\begin{aligned}
& a_2-a_1=-1-(-1)=-1+1=0 \\
& a_3-a_2=-1-(-1)=-1+1=0 \\
& a_4-a_3=-1-(-1)=-1+1=0
\end{aligned}
$
Here, the difference between the successive terms is the same, hence it is an AP.
ii)
Answer. [No]
Solution.

The given series is $0,2,0,2, \ldots$ here $a_1=0, a_2=2, a_3=0, a_4=2$

$
\begin{gathered}
a_2-a_1=2-0=2 \\
a_3-a_2=0-2=-2 \\
a_4-a_3=2-0=2
\end{gathered}
$
Here the difference in successive terms is not the same. He, once it is not an AP.
iii)
Answer. [No]
Solution.

The given series is $1,1,2,2,3,3, \ldots$ here $a_1=1, a_2=1, a_3=2, a_4=2$

$
\begin{aligned}
& a_2-a_1=1-1=0 \\
& a_3-a_2=2-1=1 \\
& a_4-a_3=2-2=0
\end{aligned}
$
Here the difference in successive terms is not the same. He, it is not an AP.
iv)
Answer. [Yes]
Solution.

The given series is $11,22,33, \ldots$ here $a_1=11, a_2=22, a_3=33$

$
\begin{aligned}
& a_2-a_1=22-11=11 \\
& a_3-a_2=33-22=11
\end{aligned}
$
Here, the difference between the successive terms is the same. Hence, it is an AP.
v)
Answer. [No]
Solution.

The given series is $\frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$

$
\begin{gathered}
a_1=\frac{1}{2}, \quad a_2=\frac{1}{3}, \quad a_3=\frac{1}{4} \\
a_2-a_1=\frac{1}{3}-\frac{1}{2}=\frac{2-3}{6}=\frac{-1}{6} \\
a_3-a_2=\frac{1}{4}-\frac{1}{3}=\frac{3-4}{12}=\frac{-1}{12}
\end{gathered}
$
Here the difference between the successive terms, is not the same. Hence, it is not an AP.
vi)
Answer. [No]
Solution.

The given series is $2,2^2, 2^3, 2^4, \ldots$

$
\begin{gathered}
a_1=2, \quad a_2=2^2=4, \quad a_3=2^3=8, \quad a_4=2^4=16 \\
a_2-a_1=4-2=2 \\
a_3-a_2=8-4=4 \\
a_4-a_3=16-8=8
\end{gathered}
$
Here the difference of the successive terms is not the same. Hence, it is not an AP.
vii)
Answer. [Yes]
Solution.

Here the given series is $\sqrt{3}, \sqrt{12}, \sqrt{27}, \sqrt{48}, \ldots$. here $\mathrm{a}_1=\sqrt{3}, \mathrm{a}_2=\sqrt{12}, \mathrm{a}_3=\sqrt{27}$,

$
\begin{aligned}
& a_4=\sqrt{48} \\
& a_2-a_1=\sqrt{12}-\sqrt{3}=2 \sqrt{3}-\sqrt{3}=\sqrt{3}=\sqrt{3}(2-1)=\sqrt{3} \\
& a_3-a_2=\sqrt{27}-\sqrt{12}=3 \sqrt{3}-2 \sqrt{3}=\sqrt{3} \\
& a_4-a_3=\sqrt{48}-\sqrt{27}=4 \sqrt{3}-3 \sqrt{3}=\sqrt{3}
\end{aligned}
$
Here the difference of the successive term is the same. Hence, it is an AP.

Question:2

Justify whether it is true to say that $-1,-\frac{3}{2},-2, \frac{5}{2}, \ldots$ forms an A.P.as $a_2 - a_1 = a_3 - a_2.$

Answer. [True]

Solution.

$
\begin{gathered}
-1,-\frac{3}{2},-2, \frac{5}{2}, \ldots \ldots \\
\text { here } \quad \mathrm{a}_1=-1, \quad \mathrm{a}_2=-\frac{3}{2}, \quad \mathrm{a}_3=-2, \quad \mathrm{a}_4=\frac{5}{2} \\
a_2-a_1=-\frac{3}{2}-(-1)=\frac{-3}{2}+1=\frac{-3+2}{2}=\frac{-1}{2} \\
a_3-a_2=-2-\left(\frac{-3}{2}\right)=-2+\frac{3}{2}=\frac{-4+3}{2}=\frac{-1}{2}
\end{gathered}
$

Here $a_2-a_1=a_3-a_2$, Hence it is an A.P.

Question:3

For the AP: –3, –7, –11, ..., can we find directly $a_{30} - a_{20}$ without actually finding $a_{30} $ and $ a_{20}$?
Give reasons for your answer.

Answer. [True]
Solution. Here, the given series is
$-3,-7,-11 \ldots$
$a_{1}=-3, a_{2}=-7$
$\begin{gathered}d=a_2-a_1 \\ =-7-(3) \\ =-7+3 \\ =-4 \\ a_{30}-a_{20}=a+(30-1) d-(a+(20-1) d) \\ =a+29 d-a-19 d \\ =10 d \\ =10 \times(-4) \\ =-40\end{gathered}$
$\text{Yes we can find}$ $\mathrm{a}_{30}-\mathrm{a}_{20}$ $\text{without actually calculating}$ $\mathrm{a}_{30} \ and \ \mathrm{a}_{20} .$

Question:4

Two APs have the same common difference. The first term of one AP is 2 and that of the other is 7. The difference between their 10th terms is the same as the difference between their 21st terms, which is the same as the difference between any two corresponding terms. Why?

Answer. [True]
Solution.
It is given that the first term of an AP is 2 and the other is 7.

Let

$
a_1=2, \quad b_1=7
$

Common difference of both A.P is same $a_{\mathrm{n}}$ is nth term for first A.P,$\quad b_n$ is $n$th term of second A.P

As per the question

$
\begin{gathered}
a_{10}-b_{10}=a_1+9 d-b_1-9 d \\
=a_1-b_1=2-7=-5 \\
a_{21}-b_{21}=a_1+20 d-b_1-20 d \\
=a_1-b_1=-5
\end{gathered}
$

Difference between first term $=a_1-b_1=-5$

It is because when we find the difference between the 10th and 21st terms, it is equal to the a₁-b_1, which is the difference between the first terms.

Question:5

Is 0 a term of the AP: 31, 28, 25, ...? Justify your answer.

Answer. [False]
Solution.
Here the given series is $31,28,25, \ldots \ldots .$.

$
\begin{aligned}
& a=31 \\
& d=a_2-a_1=28-31=-3
\end{aligned}
$
If 0 is a term of this A.P., then the value of $n$ must be a positive integer.

$
\begin{aligned}
& a_{\mathrm{n}}=a+(n-1) d \\
& 0=a+(n-1) d \\
& 0=31+(n-1)(-3) \\
& 0=31-3 n+3 \\
& 3 n=34 \\
& n=\frac{34}{3}=11.333
\end{aligned}
$
Which is not an integer.
Hence 0 is not a term of this AP.

Question:6

The taxi fare after each km, when the fare is Rs 15 for the first km and s. 8 for each additional km, does not form an AP as the total fare (in Rs) after each km is
15, 8, 8, 8, ...
Is the statement true? Give reasons.

Answer: [True]
Solution.

Given series is $15,8,8,8, \ldots$.

Here $a_1=15, \quad a_2=8, \quad a_3=8, \quad a_4=8$

$\begin{gathered}a_2-a_1=8-15=-7 \\ a_3-a_2=8-8=0 \\ a_4-a_3=8-8=0\end{gathered}$

The difference of the successive terms is not the same, hence it does not form an AP.

Therefore given statement is true.

Question:7

i)In which of the following situations do the lists of numbers involved form an AP? Give reasons for your answers.
The fee is charged to a student every month by a school for the whole session, when the monthly fee is Rs 400.
ii)
In which of the following situations do the lists of numbers involved form an AP? Give reasons for your answers.
The fee is charged every month by a school from Classes I to XI, when the monthly fee for Class I is Rs 250, and it increases by Rs 50 for the next higher class.
iii)
In which of the following situations do the lists of numbers involved form an AP? Give reasons for your answers.
The amount of money in the account of Varun at the end of every year when Rs 1000 with a simple interest of 10% per annum.
iv)
In which of the following situations do the lists of numbers involved form an AP? Give reasons for your answers.
The number of bacteria in a certain food item after each second, when they double every second.

i)

Answer. [Yes]
Solution.
According to the question:
The monthly fee charged by students is $=400$ Rs.
Hence series is: 400, 400, 400, 400 $\qquad$
The difference of the successive terms is the same, so it forms an AP with common difference $\mathrm{d}=400-400=0$
ii)

Answer. [Yes]
Solution.
According to the question:
Fees started at Rs. 250 and increased by 50 Rs. from I to XII are

Here first term $\left(a_1\right)=250$

common difference

$
\begin{aligned}
& (d)=50, \quad a_1=250 \\
& a_2=250+50=300 \\
& a_3=300+50=350 \\
& a_4=350+50=400
\end{aligned}
$

So the A.P. is:

$
250,300,350,400 \ldots
$

It forms an AP with a common difference of 50.

iii)

Answer. [Yes]
Solution. We know that: Simple interest $=(P \times R \times T) / 100$

This is definite from an AP with a common difference of $\mathrm{SI}=100$ Rs.
iv)

Answer. [No]
Solution.
Let the number of bacteria $=\mathrm{x}$
According to the question:
They double in every second.
Here $a_1=x, \quad a_2=2 x, \quad a_3=4 x, \quad a_4=8 x, \ldots x, 2 x, 4 x, 8 x, \ldots$
Since the difference between the successive terms is not the same therefore it does not form an AP.

Question:8

i)
Justify whether it is true to say that the following are the nth terms of an AP.
2n–3
ii)
Justify whether it is true to say that the following are the nth terms of an AP.
$3n^2+5$
iii)
Justify whether it is true to say that the following are the nth terms of an AP.
$1+n+n^2$

i)

Answer: [Yes]
Solution.

Hence, $2n-3$ is the $ n$th term of an AP because the common difference is the same.
ii)

Answer. [ No ]
Solution.

Here, $a_n=3 n^2+5$
Put $n=1, \quad a_1=3(1)^2+5=3+5=8$
Put $n=2, \quad a_2=3(2)^2+5=12+5=17$
Put $n=3, \quad a_3=3(3)^2+5=27+5=32$
Numbers are:

$
\begin{gathered}
8,17,32 \ldots \\
a_2-a_1=17-8=9 \\
a_3-a_2=32-17=15
\end{gathered}
$
Here, the common difference is not the same
Therefore series $ 3n^2+5$ is not the nth term of an AP.
iii)

Answer. [No]
Solution.

$
\text { a }_{\mathrm{n}}=1+n+n^2 P u t n=1, a_1=1+1+(1)^2=3 P u t n=2, a_2=1+2+(2)^2=7 P u t n=3, a_3=1+3+(3)^2=13 \text { Thenumbersar c } 3,7,13 \ldots a_2-a_1=7-3=4 a_3-a_2=13-7=6
$

Here, the common difference is not the same.

Question:1

Match the APs given in column A with the suitable common differences given in column B.

Answer:

$\begin{aligned} & \left(A_1\right) \\ & 2,-2,-6,-10, \ldots \\ & a_2=-2, a_1=2 \\ & d=a_2-a_1 \\ & -2-2=-4 \quad\left(B_4\right)-4 \\ & \left(A_2\right) \\ & a=-18, n=10, a_{\mathrm{n}}=0 \\ & a_{\mathrm{n}}=0 \\ & \because a+(n-1) d=0 \\ & -18+(10-1) d=0 \\ & 9 d=18 \\ & d=2 \quad\left(B_5\right) 2 \\ & \begin{array}{l}\left(A_3\right) \\ a=0, a_{10}=6 \\ a+9 d=6 \\ 0+9 d=6 \\ d=\frac{6}{9}=\frac{2}{3} \quad\left(B_1\right) 2 / 3 \\ \left(A_4\right) \\ a_2=13, a_4=3 \\ \because a+(n-1) d=0 \\ a+d=13 \\ a+3 d=3 \\ -------- \\ -2 d=10 \\ d=\frac{10}{-2} \\ d=-5 \quad\left(B_2\right)-5\end{array}\end{aligned}$

Question:2

i)

Verify that each of the following is an AP, and then write its next three terms.
$0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, \ldots$
ii)
Verify that each of the following is an AP, and then write its next three terms.
$5, \frac{14}{3}, \frac{13}{3}, 4, \ldots \ldots$
iii)
Verify that each of the following is an AP, and then write its next three terms.
$\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, \ldots$
iv)
Verify that each of the following is an AP, and then write its next three terms.
a + b, (a + 1) + b, (a + 1) + (b + 1), ...
v)
Verify that each of the following is an AP, and then write its next three terms.
a, 2a + 1, 3a + 2, 4a + 3,...

Answer:

i)

Here is the given series,

$
\begin{aligned}
& 0, \frac{1}{4}, \frac{1}{2}, \frac{3}{4} \\
& a_1=0, a_2=\frac{1}{4} a_3=\frac{3}{4} \\
& a_2-a_1=\frac{1}{4}-0=\frac{1}{4} \\
& a_3-a_2=\frac{1}{4} \\
& a_4-a_3=\frac{1}{4}
\end{aligned}
$
Here, the common difference is the same. Hence given series is an AP

$
\begin{aligned}
& a_5=a_1+4 d\left(u \operatorname{sing} a_n=a_1+(n-1) \cdot d\right) \\
& a_6=a_1+5 d \\
& a_7=a_1+6 d \\
& =0+6\left(\frac{1}{4}\right) \\
& =\frac{3}{2}
\end{aligned}
$
The next three terms are

$
1, \frac{5}{4}, \frac{3}{2}
$

ii)

Here the given series is, $5, \frac{14}{3}, \frac{13}{3}, 4, \ldots \ldots$
Here, the common difference is the same.
Hence, the given series is an AP.

$
\begin{aligned}
& a_5=a_1+4 d\left(u \operatorname{sing} a_{\mathrm{n}}=a_1+(n-1) \cdot d\right) \\
& a_6=a_1+5 d \\
& a_7=a_1+6 d
\end{aligned}
$
Next terms are $\frac{11}{3}, \frac{10}{3}, 3$

iii)

Here the given series is $\sqrt{3}, 2 \sqrt{3}, 3 \sqrt{3}, \ldots$
Here the common difference is the same. Hence, the given series is an AP.

$
\begin{gathered}
a_4=a_1+3 d \quad\left(\text { Using } a_n=a_1+(n-1) \cdot d\right) \\
=\sqrt{3}+3 \sqrt{3}=4 \sqrt{3} \\
\\
a_5=a_1+4 d \\
=\sqrt{3}+4 \sqrt{3}=5 \sqrt{3} \\
\\
a_6=a_1+5 d \\
= \\
\sqrt{3}+5 \sqrt{3}=6 \sqrt{3}
\end{gathered}
$

Next terms are:

$
4 \sqrt{3}, 5 \sqrt{3}, 6 \sqrt{3}
$
iv)

Here the given series is $a+b,(a+1)+b,(a+1)+(b+1), \ldots$

$
\begin{gathered}
a_1=a+b, \quad a_2=(a+1)+b, \quad a_3=(a+1)+(b+1) \\
a_2-a_1=(a+1)+b-(a+b)=1 \\
a_3-a_2=(a+1)+(b+1)-[(a+1)+b]=1
\end{gathered}
$

Here the common difference is the same, hence the given series is an AP.

$
\begin{gathered}
a_4=a_1+3 d \quad\left(\text { Using } a_n=a_1+(n-1) \cdot d\right) \\
=(a+b)+3(1)=(a+1)+(b+1)+1 \\
a_5=a_1+4 d \\
=(a+b)+4(1)=(a+1)+(b+1)+1+1 \\
a_6=a_1+5 d \\
=(a+b)+5(1)=(a+1)+(b+1)+1+1+1
\end{gathered}
$

Next three terms are:

$
(a+1)+(b+1)+1, \quad(a+1)+(b+1)+1+1, \quad(a+1)+(b+1)+1+1+1
$

v)

Here the given series is $a, 2 a+1,3 a+2,4 a+3, \ldots$

$
\begin{gathered}
a_1=a, \quad a_2=2 a+1, \quad a_3=3 a+2 \\
a_2-a_1=(2 a+1)-a=a+1 \\
a_3-a_2=(3 a+2)-(2 a+1)=a+1
\end{gathered}
$

Here the common difference is the same, hence the given series is an AP.

$
\begin{aligned}
& a_5=a_1+4 d \quad\left(\text { Using } a_n=a_1+(n-1) \cdot d\right) \\
& =a+4(a+1)=a+4 a+4=5 a+4 \\
& =a+5(a+1)=a+5 a+5=6 a+5 \\
& a_6=a_1+5 d \\
& a_7=a_1+6 d \\
& =a+6(a+1)=a+6 a+6=7 a+6
\end{aligned}
$

Next three terms are:

$
5 a+4, \quad 6 a+5, \quad 7 a+6
$

Question:3

i)
Write the first three terms of the APs when a and d are as given below:
$\mathrm{a}=\frac{1}{2}, \mathrm{~d}=-\frac{1}{6}$
ii)
Write the first three terms of the APs when a and d are as given below:
a = – 5, d = – 3
iii)
Write the first three terms of the APs when a and d are as given below:
$\mathrm{a}=\sqrt{2}, \mathrm{~d}=\frac{1}{\sqrt{2}}$

Answer:

i)

$
\begin{gathered}
\text { Here } a=\frac{1}{2}, \quad d=-\frac{1}{6} \\
a_1=a=\frac{1}{2} \\
a_2=a+d \quad\left(\text { Using } a_n=a+(n-1) \cdot d\right) \\
a_3=a+2 d \\
=\frac{1}{2}+2\left(\frac{-1}{6}\right) \\
=\frac{1}{2}-\frac{2}{6} \\
=\frac{3-2}{6}=\frac{1}{6}
\end{gathered}
$

Hence, the first three terms are:

$
\frac{1}{2}, \quad \frac{1}{3}, \quad \frac{1}{6}, \ldots
$
ii)

Here $a=-5, \quad d=-3$

$
\begin{gathered}
a_1=a=-5 \\
a_2=a+d \quad\left(\mathrm{Using} a_n=a+(n-1) \cdot d\right) \\
=-5-3=-8 \\
\\
a_3=a+2 d \\
=-5+2(-3) \\
=
\end{gathered}
$

Here, the first three terms are:

$
-5, \quad-8, \quad-11 .
$
iii)

Here $a=\sqrt{2}, \quad d=\frac{1}{\sqrt{2}}$

$
\begin{aligned}
a_2=a+d \quad & \left(\text { Using } a_n=a+(n-1) d\right) \\
& =\sqrt{2}+\frac{1}{\sqrt{2}} \\
= & \frac{2+1}{\sqrt{2}}=\frac{3}{\sqrt{2}} \\
& a_3=a+2 d \\
= & \sqrt{2}+2\left(\frac{1}{\sqrt{2}}\right) \\
& =\sqrt{2}+\frac{2}{\sqrt{2}} \\
= & \frac{2+2}{\sqrt{2}}=\frac{4}{\sqrt{2}}
\end{aligned}
$

Here, the first three terms are:

$
\sqrt{2}, \quad \frac{3}{\sqrt{2}}, \quad \frac{4}{\sqrt{2}}
$

Question:4

Find a, b and c such that the following numbers are in AP: a, 7, b, 23, c.

Answer:

Given AP is a, $7, b, 23, c$
We know that an f series is in AP, then

$
a_1=a, a_2=7, a_3=b, a_4=23, a_5=c
$

(We know that the common difference of an A.P. is Equal)

$
\begin{aligned}
& a_2-a_1=a_3-a_2=a_4-a_3=a_5-a_4 \\
& \therefore 7-a=b-7=23-b=c-23
\end{aligned}
$
Taking $2^{\text {nd }}$ and $3^{\text {rd }}$ term

$
\begin{aligned}
& b-7=23-b \\
& 2 b=30 \\
& b=\frac{30}{2}=15
\end{aligned}
$
Taking $1^{\text {st }}$ and $2^{\text {nd }}$ term

$
\begin{aligned}
& 7-a=b-7 \\
& 7-a=15-7(\because b=15) \\
& 7-8=a \\
& -1=a
\end{aligned}
$
Taking $1^{\text {st }}$ and fourth term

$
\begin{aligned}
& 7-a=c-23 \\
& 7-(-1)=c-23(\because a=-1) \\
& 7+1=c-23 \\
& 8+23=6 \\
& 31=c
\end{aligned}
$
Hence $a=-1, b=15, c=31$

Question:5

Determine the AP whose fifth term is 19 and the difference between the eighth term from the thirteenth term is 20.

Answer:

Given:
The fifth term is 19 :

$
a+4 d=19
$

The difference between the eighth and thirteenth terms is 20 :

$
(a+12 d)-(a+7 d)=20
$

This simplifies to $ 5d=20$, so $d=4$.

Substitute $d=4$ into the first equation:

$
a+16=19, \text { so } a=3
$

Therefore, the first term is $a=3$ and the common difference is $d=4$.

Question:6

The 26th, 11thand the last term of an AP are 0, 3 and -1/5, respectively. Find the common difference and the number of terms.

Answer:

$
\text { Given } \mathrm{a}_{26}=0, a_{11}=3, a_{\mathrm{n}}=-\frac{1}{5}
$
$
\begin{aligned}
& a+25 d=0 \ldots(1)\left(u \operatorname{sing} \mathrm{a}_{\mathrm{n}}=a+(n-1) d\right) \\
& a+10 d+3 \ldots(2)\left(u \operatorname{sing} a_{\mathrm{n}}=a+(n-1) d\right) \\
& 15 d=-3
\end{aligned}
$
Put $d=-\frac{1}{5}$ in (1) we get

$
\begin{aligned}
& a-5=0 \\
& a=5 \\
& a_{\mathrm{n}}=-\frac{1}{5}
\end{aligned}
$
We know that $\mathrm{a}_{\mathrm{n}}=a+(n-1) d$

$
\begin{aligned}
& \Rightarrow-\frac{1}{5}-5=(n-1)\left(-\frac{1}{5}\right) \\
& \Rightarrow \frac{-1-25}{5}=(n-1)\left(-\frac{1}{5}\right) \\
& \Rightarrow \frac{-26}{5}=(n-1)\left(-\frac{1}{5}\right) \\
& \Rightarrow \frac{-26}{5} \times \frac{-5}{1}=n-1 \\
& \therefore 26=n-1 \\
& \Rightarrow n=26+1 \\
& \Rightarrow n=27
\end{aligned}
$
ANS: -

$
d=-\frac{1}{5}, n=27
$

Question:7

The sum of the 5th and the 7th terms of an AP is 52, and the 10th term is 46. Find the AP.

Given $\mathrm{a}_5+\mathrm{a}_7=52$

$
a_5=a+4 d a_7=a+6 d\left(u \sin g a_{\mathrm{n}}=a+(n-1) \cdot d\right)
$
Given $\mathrm{a}_{10}=46$

$
\begin{aligned}
& \therefore a+9 d=46 \ldots(1) \\
& a+4 d+a+6 d=52 \\
& 2 a+10 d=52
\end{aligned}
$
Dividing both sides by 2

$
\begin{aligned}
& a+5 d=26 \ldots(2) \\
& a+9 d=-46 \\
& -4 d=-20 \\
& d=\frac{-20}{-4} \\
& d=5
\end{aligned}
$
Put $\mathrm{d}=5$ in equation (1) we get

$
\begin{aligned}
& a+5 \times(5)=26 \\
& a+25=26 \\
& a=26-25 \\
& a=1 \\
& a_1=a=1 \\
& a_2=a+d \\
& =1+5 \\
& =6 \\
& a_3=a+2 d \\
& =1+2 \times(5) \\
& =1+10 \\
& =11
\end{aligned}
$
Hence the AP is $ 1,6,11\ldots$.

Question:8

Find the 20th term of the AP whose 7th term is 24 less than the 11th term, the first term being 12.

Given $\mathrm{a}=12$

$
\begin{aligned}
& a_7=a_{11}-24 \\
& a+6 d=a+10 d-24\left(\text { using } \mathrm{a}_{\mathrm{n}}=a+(n-1) \cdot d\right) \\
& a-a+24=10 d-6 d \\
& 24=4 d \\
& d=6
\end{aligned}
$
We know that

$
a_{\mathrm{n}}=a+(n-1) d
$
$
\begin{aligned}
& \text { Put } \mathrm{n}=20, a_{20}=a+(20-1) d \\
& =12+(19) \times(6) \\
& =12+114 \\
& =126
\end{aligned}
$
Hence

$
20^{\text {th }} \text { term is } 126
$

Question:9

If the 9th term of an AP is zero, prove that its 29th term is twice its 19th term.

Answer:

Let the first term = a
common difference = d
Given:- $a_{\text{9}} = 0\\$

$
\begin{aligned}
\Rightarrow \therefore a+8 d & =0 \quad\left(\text { using } a_n=a+(n-1) d\right) \\
& \Rightarrow \therefore a=-8 d \ldots(1)
\end{aligned}
$

To prove:

$
\begin{gathered}
2 a_{19}=a_{29} \\
a_{19}=a+18 d \quad\left(\text { using } a_n=a+(n-1) d\right) \\
=-8 d+18 d \quad\{a=-8 d \text { from equation }(1)\} \\
=10 d \ldots(2) \\
a_{29}=a+(29-1) d \\
=a+28 d \\
=-8 d+28 d \quad\{a=-8 d \text { from equation }(1)\} \\
=20 d \\
=2 \times 10 d
\end{gathered}
$

By equation (2),

$
a_{29}=2 a_{19}
$
Hence proved

Question:10

Find whether 55 is a term of the AP: 7, 10, 13, ______or not. If yes, find which term it is.

Answer:

Let 55 be the nth term of an AP
$\therefore a_{\text{n}} = 55 $ \ldots $ (1)\\$
Given AP is 7, 10, 13........
Here a = 7, d = 10 - 7
d = 3
By equation (1) is $a_{\text{n}} = 55\\$
$\begin{gathered}a+(n-1) \times d=55 \quad \ldots(2) \quad\left(\text { using } a_n=a+(n-1) d\right) \\ 7+(n-1) \times 3=55 \\ (n-1) \times 3=55-7 \\ (n-1) \times 3=48 \\ n-1=\frac{48}{3} \\ n-1=16 \\ n=17\end{gathered}$
Here n is a positive integer.
Hence, 55 is the 17^th term of an AP

Question:11

Determine k so that $k^2+ 4k + 8, 2k^2 + 3k + 6, 3k^2 + 4k + 4$ are three consecutive terms of an AP.

Answer:

Here, the given AP is

$
\begin{gathered}
k_2+4 k+8, \quad 2 k_2+3 k+6, \quad 3 k_2+4 k+4 \\
a_1=k_2+4 k+8, \quad a_2=2 k_2+3 k+6, \quad a_3=3 k_2+4 k+4 \\
a_2-a_1=2 k_2+3 k+6-\left(k_2+4 k+8\right) \\
=2 k_2+3 k+6-k_2-4 k-8 \\
=k_2-k-2 \\
a_3-a_2=3 k_2+4 k+4-\left(2 k_2+3 k+6\right) \\
=3 k_2+4 k+4-2 k_2-3 k-6 \\
=k_2+k-2
\end{gathered}
$

Hence, it is given that these terms are an AP
So the common difference between the A.P

$
\begin{gathered}
a_2-a_1=a_3-a_2 \\
k_2-k-2=k_2+k-2 \\
k_2-k-2-k_2-k+2=0 \\
-2 k=0 \\
k=0
\end{gathered}
$

Question:12

Split 207 into three parts such that these are in AP and the product of the two smaller parts is 4623.

Answer:

Let the three parts be (a-d), a, (a+d)
As per the question -

$
\begin{gathered}
a-d+a+a+d=207 \\
3 a=207 \\
a=\frac{207}{3}=69 \\
a=69
\end{gathered}
$

It is given that the product of the two similar parts is 4623

$
\begin{gathered}
(a-d) \times a=4623 \\
(69-d) \times 69=4623 \\
69-67=d \\
d=2
\end{gathered}
$

So the required terms are

$
\begin{gathered}
a-d=69-2=67 \\
a=69 \\
a+d=69+2=71
\end{gathered}
$
The required numbers are 67, 69, and 71

Question:13

The angles of a triangle are in AP. The greatest angle is twice the least. Find all the angles of the triangle.

Answer:

Let the three angles $\mathrm{be}(a-d), a,(a+d)$ in AP.
According to the question, the greatest is twice the least:

$
\begin{gathered}
(a+d)=2(a-d) \\
a+d=2 a-2 d \\
2 a-a-2 d-d=0 \\
a-3 d=0 \\
a=3 d
\end{gathered}
$

We know that the sum of the angles of a triangle is $180^{\circ}$ :

$
\begin{gathered}
a+a-d+a+d=180^{\circ} \\
3 a=180^{\circ} \\
a=\frac{180^{\circ}}{3}=60^{\circ}
\end{gathered}
$

Substituting $a=60^{\circ}$ in $a=3 d$ :

$
\begin{gathered}
60^{\circ}=3 d \\
d=20^{\circ}
\end{gathered}
$

So the angles are:

$
\begin{gathered}
a-d=60^{\circ}-20^{\circ}=40^{\circ} \\
a=60^{\circ} \\
a+d=60^{\circ}+20^{\circ}=80^{\circ}
\end{gathered}
$

Required angles: $40^{\circ}, 60^{\circ}, 80^{\circ}$

Question:14

If the nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same, find the value of n. Also, find that term.

Answer:

The given APs are

$
9,7,5, \ldots \text { and } 24,21,18, \ldots
$
In AP

$
9,7,5, \ldots \ldots
$

$
\begin{aligned}
& a_1=a=9, \quad a_2=7 \\
& d_1=a_2-a_1=7-9=-2 \\
& a_n=a+(n-1) d_1
\end{aligned}
$

(1) In the AP $24,21,18, \ldots$

$
\begin{aligned}
& b_1=b=24, \quad b_2=21 \\
& d_2=b_2-b_1=21-24=-3 \\
& b_n=b+(n-1) d_2
\end{aligned}
$

Here, the number of terms $n$ is equal in both cases
According to the question

$
a_{\mathrm{n}}=b_{\mathrm{n}}
$

Question:15

If the sum of the 3rd and the 8th terms of an AP is 7 and the sum of the 7th and the 14th terms is 3, find the 10th term.

Answer:

It is given that the sum of $3^{\text {rd }}$ and $8^{\text {th }}$ term is $7 \cdot$

$
\begin{aligned}
& \text { Solve }(1) \text { and }(2) 2 a+19 d=-3 \\
& 2 a+9 d=7 \\
& 10 d=-10 \\
& d=-1 \\
& \text { Putd }=-1 \text { in }(1) \\
& \Rightarrow 2 a+9(-1)=7 \\
& \Rightarrow 2 a=7+9 \\
& \Rightarrow a=\frac{16}{2}=8 \\
& a_{10}=(a+(10-1) d) \\
& \Rightarrow a_{10}=8+9(-1) \\
& \Rightarrow a_{10}=8-9 \\
& \Rightarrow a_{10}=-1
\end{aligned}
$

Question:16

Find the 12th term from the end of the AP: -2, -4, -6,..., -100.

Answer:

The given AP : is $-2,-4,-6, \ldots \ldots .,-100$
AP from end is $-100,-98 \ldots \ldots . . . .4,-6,-2$

$
\begin{aligned}
& a_1=a=-100 \\
& d=a_2-a_1=-98-(-100)=-98+100=2 \\
& a_{12}=a+(12-1) \cdot d\left(\text { using } a_{\mathrm{n}}=a+(n-1) \times d\right) \\
& a_{12}=-100+11 \times(2) \\
& a_{12}=-100+22 \\
& a_{12}=-78
\end{aligned}
$

Question:17

Which term of the AP: 53, 48, 43,... is the first negative term?
Answer:

The given arithmetic progression (AP) is:
$53,48,43, \ldots$
Here, the first term $a=53$ and the common difference $d=48-53=-5$.
To find the first negative term, we use the formula for the $n$-th term of an AP:

$
T_n=a+(n-1) \cdot d
$

We want to find $n$ such that $T_n<0$.
Substitute $a=53$ and $d=-5$ :

$
\begin{gathered}
T_n=53+(n-1)(-5) \\
T_n=53-5(n-1) \\
T_n=53-5 n+5 \\
T_n=58-5 n
\end{gathered}
$

Set $T_n<0$ to find the first negative term:

$
\begin{gathered}
58-5 n<0 \\
58<5 n \\
n>\frac{58}{5}=11.6
\end{gathered}
$

Since $n$ must be a whole number, the first negative term occurs at $n=12$. Thus, the first negative term is the 12th term of the AP.

Question:18

How many numbers lie between 10 and 300, which when divided by 4 leave a remainder of 3?

Answer:

The numbers between 10 and 300 which when divided by 4 leave remainder 3 are: $11,15,19,23, \ldots, 299$

$
\begin{gathered}
a_1=11, \quad a_2=15, \quad a_3=19 \\
a_2-a_1=15-11=4 \\
a_3-a_2=19-15=4
\end{gathered}
$

It is an AP with $a_1=a=11, \quad d=4$
Let there be $n$ terms in this AP

$
\begin{gathered}
a_{\mathrm{n}}=299 \\
a+(n-1) \times d=299 \quad\left(\text { using } a_{\mathrm{n}}=a+(n-1) d\right) \\
11+(n-1) \times 4=299 \\
(n-1) \times 4=299-11 \\
(n-1) \times 4=288 \\
(n-1)=\frac{288}{4} \\
(n-1)=72 \\
n=72+1 \\
n=73
\end{gathered}
$

Question:19

Find the sum of the two middlemost terms of the AP:
$-\frac{4}{3},-1,-\frac{2}{3}, \ldots, 4 \frac{1}{3}$

Answer:

Here, the given AP is $-\frac{4}{3},-1,-\frac{2}{3}, \ldots, 4 \frac{1}{3}$
In this AP: $a_1=a=-\frac{4}{3}$

$
d=a_2-a_1=-1-\left(-\frac{4}{3}\right)=-1+\frac{4}{3}=\frac{-3+4}{3}=\frac{1}{3}
$

Let there be $n$ terms in the AP
The nth term of AP is $4 \frac{1}{3}$

$
\begin{gathered}
a_n=4 \frac{1}{3} \\
a+(n-1) \cdot d=\frac{13}{3} \quad\left(\mathrm{Using} a_n=a+(n-1) d\right) \\
-\frac{4}{3}+(n-1) \times\left(\frac{1}{3}\right)=\frac{13}{3} \\
(n-1) \times\left(\frac{1}{3}\right)=\frac{13+4}{3} \\
n-1=17 \\
n=18
\end{gathered}
$

The total terms in this AP are 18, and the middlemost terms are $a_9$ and $a_{10}$

$
\begin{gathered}
a_9+a_{10}=a+(9-1) d+a+(10-1) d \quad\left(\text { using } a_n=a+(n-1) d\right) \\
=2 a+17 d \\
=2\left(-\frac{4}{3}\right)+17\left(\frac{1}{3}\right) \\
=\frac{-8}{3}+\frac{17}{3} \\
a_9+a_{10}=\frac{-8+17}{3}=\frac{9}{3}=3 \\
a_9+a_{10}=3
\end{gathered}
$

Question:20

The first term of an AP is -5 and the last term is 45. If the sum of the terms of the AP is 120, then find the number of terms and the common difference.

Answer:

Given that $a=-5, \quad a_n=45, \quad S_n=120$

$
a_n=45
$

$
\begin{gathered}
a+(n-1) \times d=45 \quad\left(\text { Using } a_n=a+(n-1) d\right) \\
-5+(n-1) \times d=45 \\
(n-1) \times d=50 \quad \ldots \quad(1) \\
S_n=120 \\
S_n=\frac{n}{2}[2 a+(n-1) d]=120 \\
\frac{n}{2}[-10+50]=120 \quad(\text { Using }(1)) \\
\frac{n}{2} \times 40=120 \\
n \times 20=120 \\
n=\frac{120}{20}=6
\end{gathered}
$

Number of terms $n=6$
Put $n=6$ in (1)

$
\begin{gathered}
(6-1) \times d=50 \\
5 d=50 \\
d=10
\end{gathered}
$

Common difference $d=10$

Question:21

i)Find the sum:1 + (-2) + (-5) + (-8) + ... + (-236)
ii)Find the sum:$4-\frac{1}{n}+4-\frac{2}{n}+4-\frac{3}{n}+\ldots$ up to n terms
iii) Find the sum:$\frac{a-b}{a+b}+\frac{3 a-2 b}{a+b}+\frac{5 a-3 b}{a+b}+.$ to 11 terms

Answer:

i)

The given series is $1+(-2)+(-5)+(-8)+\cdots+(-236)$
Here, $a_1=a=1, \quad a_2=-2$

$
\begin{gathered}
d=a_2-a_1=-2-1=-3 \\
a_n=-236
\end{gathered}
$

$
\begin{gathered}
a+(n-1) \times d=-236 \quad\left(\text { Using } a_n=a+(n-1) d\right) \\
1+(n-1) \times(-3)=-236 \\
(n-1) \times(-3)=-237 \\
n-1=79 \\
n=80
\end{gathered}
$

The formula for the sum is:

$
\begin{gathered}
S_n=\frac{n}{2}[2 a+(n-1) d] \\
S_n=\frac{80}{2}[2 \times 1+(80-1) \times(-3)] \\
=40[2+79 \times(-3)] \\
=40[2-237] \\
=40 \times(-235) \\
S_n=-9400
\end{gathered}
$
ii)

The given series is $\left[4-\frac{1}{n}\right]+\left[4-\frac{2}{n}\right]+\left[4-\frac{3}{n}\right]+\ldots$

$
\begin{gathered}
\text { First term }(a)=\left[4-\frac{1}{n}\right] \\
\begin{array}{c}
d=\left[4-\frac{2}{n}\right]-\left[4-\frac{1}{n}\right] \\
=\frac{-2}{n}+\frac{1}{n} \\
=\frac{-1}{n} \\
S_n=\frac{n}{2}\left[2\left(4-\frac{1}{n}\right)+(n-1) \times\left(\frac{-1}{n}\right)\right] \\
=\frac{n}{2}\left[8-\frac{2}{n}-\frac{n-1}{n}\right] \\
= \\
\frac{n}{2}\left[8-\frac{2+n-1}{n}\right] \\
= \\
\frac{n}{2}\left[8-\frac{n+1}{n}\right] \\
= \\
\frac{n}{2} \times \frac{8 n-(n+1)}{n} \\
=
\end{array} \\
=\frac{n}{2} \times \frac{7 n-1}{n} \\
=\frac{7 n-1}{2} \\
S_n=\frac{7 n-1}{2}
\end{gathered}
$
iii)

The given series is $\frac{a-b}{a+b}+\frac{3 a-2 b}{a+b}+\frac{5 a-3 b}{a+b}+\ldots$

$
\begin{gathered}
a_1=\frac{a-b}{a+b}, \quad a_2=\frac{3 a-2 b}{a+b} \\
d=a_2-a_1
\end{gathered}
$

Given that $n=11$

$
\begin{gathered}
S_{11}=\frac{11}{2}\left[2 a_1+(11-1) d\right] \\
=\frac{11}{2}\left[\frac{2(a-b)}{a+b}+\frac{10(2 a-b)}{a+b}\right] \\
=\frac{11}{2}\left[\frac{2 a-2 b+20 a-10 b}{a+b}\right] \\
=\frac{11}{2}\left[\frac{22 a-12 b}{a+b}\right] \\
=\frac{11}{2} \times \frac{2(11 a-6 b)}{a+b} \\
S_{11}=\frac{11(11 a-6 b)}{a+b}
\end{gathered}
$

Question:22

Which term of the AP: -2, -7, -12,... will be -77? Find the sum of this AP up to the term -77.

Answer:

The given AP is $-2,-7,-12, \ldots$
First term $(a)=-2$
Common difference $(d)=-7-(-2)=-7+2=-5$
Let the $n^{\text {th }}$ term be -77

$
\begin{gathered}
a_n=-77 \\
a+(n-1) \cdot d=-77 \\
-2+(n-1) \cdot(-5)=-77 \\
(n-1) \cdot(-5)=-77+2 \\
(n-1)=\frac{-75}{-5}=15 \\
n=15+1=16
\end{gathered}
$

Hence, the $16^{\text {th }}$ term is -77
Sum of 16 terms

$
\begin{gathered}
S_{16}=\frac{16}{2}[2(-2)+(16-1)(-5)] \\
=8[-4+15(-5)] \\
=8[-4-75] \\
S_{16}=8 \times(-79) \\
S_{16}=-632
\end{gathered}
$

Question:23

If $a_n = 3 - 4n,$ $\text{show that} a_{\text{1}}, a_{\text{2}}, a_{\text{3}} ,...$ form an AP. Also find $S_{20}$

Answer:

Here it is given that

$a_{\text{n}} = 3 - 4n\\$

Put n =1 then

$a_{\text{1}} = 3 - 4(1)\\$

$a_{\text{1}} = 3 - 4= - 1\\$

Put n =2 then

$a_{\text{2}} = 3 - 4(2)\\$

$a_{\text{2}} = 3 \textbf{-} 8 = \textbf{- }5\\$

Put n =3 then

$a_{\text{3}} = 3 - 4(3)\\$

$a_{\text{3}} = 3 - 12 = - 9\\$

$\text{Then A.P is -1, -5, -9} \ldots \ldots $

$ a_{\text{2}} - a_{\text{1}}= - 5 - (-1) = - 5 + 1 = - 4\\ a_{\text{3}} - a_{\text{2}} = - 9 - (-5) = - 9 + 5 = - 4$

$ a_{\text{2}} - a_{\text{1}} = a_{\text{3}} - a_{\text{2}}\\ \text{ Hence}, a_{\text{1}}, a_{\text{2}}, a_{\text{3}} \ldots \text{ forms an AP with } a_{1 }= a = - 1 , d = -4$

$ \text{ Here d= common difference and first term is a} = a_{\text{1}}= - 1$

$S_{2}=\frac{20}{2}\left[ 2\left( -1 \right)+\left( 20-1 \right)\left( -4 \right) \right] \left( Using \ S_{n}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] \right) \\ = 10 [- 2 + 19(-4)]$

$= 10[-2 - 76]\\ S_{20} = - 780\\$

Question:24

In an Ap if $S_n = n (4n + 1)$ find the A.P.

Answer:

We are given that the sum of the first $n$ terms of the AP is:

$
S_n=n(4 n+1)
$
The first term $a$ is $S_1=5$.
For the second term, $S_2=18$, so the second term $T_2=S_2-S_1=18-5=13$.
The common difference $d$ is:

$
d=T_2-T_1=13-5=8
$
The general term of the AP is:

$
T_n=a+(n-1) d=5+(n-1) \cdot 8=8 n-3
$
Thus, the arithmetic progression is:

$
5,13,21,29, \ldots
$
And the general term is:

$
T_n=8 n-3
$

Question:25

In an AP, if $S_{n} = 3n_{2} + 5n $ and $a_{k} = 164,$ find the value of k

Answer:

Here it is given that $S_{\mathrm{n}}=3 n^2+5 n \ldots \ldots$ (1)

$
\begin{aligned}
& a_{\mathrm{k}}=164 \\
& \text { Weknowthat } S_1=a, S_2=a+(a+d) \\
& \text { Putn }=1 \text { inequation }(1) \\
& S_1=3(1)^2+5(1)=8 \\
& \text { Putn }=2 \text { inequation }(1) \\
& S_2=3(2)^2+5(2) \\
& S_2=12+10=22 \\
& a=8 \\
& a+a+d=22 \\
& 2 a+d=22 \\
& 2(8)+d=22 \\
& d=22-16 \\
& d=6 \\
& T h e A P is a, a+d, a+2 d \ldots \\
& a=8 \\
& a+d=8+6=14 \\
& a+2 d=8+6(2)=8+12=20 \\
& 8,14,20 \ldots \\
& a_{\mathrm{k}}=164 \\
& a+(k-1) \cdot d=164 \\
& 8+(k-1) \cdot 6=164 \\
& (k-1) \cdot 6=164-8 \\
& k=26+1 \\
& k=27
\end{aligned}
$

Question:26

If $S_{n}$ denotes the sum of the first n terms of an AP, prove that $\\ S_{12} = 3(S_{8} - S_{4})\\$

Answer:

To prove : $-S_{12}=3\left(S_8-S_4\right)$
If $S_n$ is the sum of $n$ terms, then the first term is $a$ and the common difference is $d$

$
\begin{gathered}
S_n=\frac{n}{2}[2 a+(n-1) d] \quad \ldots(1) \\
\text { Put } n=12 \text { in equation (1) } \\
S_{12}=\frac{12}{2}[2 a+11 d]=6[2 a+11 d] \\
S_{12}=12 a+66 d \quad \ldots(2) \\
S_8=\frac{8}{2}[2 a+7 d]=4[2 a+7 d] \\
S_8=8 a+28 d \quad \ldots(3) \\
S_4=\frac{4}{2}[2 a+3 d]=2[2 a+3 d] \\
S_4=4 a+6 d \quad \ldots(4) \\
\text { RHS: } 3\left(S_8-S_4\right) \\
=3(8 a+28 d-4 a-6 d) \quad(\text { By Equation }(3) \text { and }(4)) \\
=3(4 a+22 d) \\
=12 a+66 d \quad(\text { By Equation }(2)) \\
\quad=S_{12}
\end{gathered}
$

Question:27

Find the sum of the first 17 terms of an AP whose 4th and 9th terms are -15 and -30, respectively.

Answer:

$
\begin{gathered}
a_4=-15, \quad a_9=-30 \\
a+3 d=-15 \quad \ldots(1) \\
a+8 d=-30 \quad \ldots(2) \\
\left(\text { Using } a_n=a+(n-1) \cdot d\right) \\
\text { From equations (1) and (2): } \\
(a+8 d)-(a+3 d)=-30+15 \\
5 d=-15 \\
d=-3
\end{gathered}
$

Put $d=-3$ in equation (1):

$
\begin{gathered}
a+3(-3)=-15 \\
a-9=-15 \\
a=-6 \\
S_{17}=\frac{17}{2}[2(-6)+(17-1)(-3)] \\
\left(\mathrm{U} \operatorname{sing} S_n=\frac{n}{2}[2 a+(n-1) d]\right) \\
=\frac{17}{2}[-12-48] \\
=\frac{17}{2} \times(-60) \\
S_{17}=-510
\end{gathered}
$

Question:28

If the sum of the first 6 terms of an AP is 36 and that of the first 16 terms is 256, find the sum of the first 10 terms.

Answer:

$\begin{aligned} & S_6=36, S_{16}=256 \\ & {\left[U \operatorname{sing} S_n=\frac{n}{2}[2 a+(n-1) d]\right]} \\ & S_6=36 S_{16}=256 \\ & =\frac{6}{2}[2 a+(6-1) d]=36 \frac{16}{2}[2 a+(16-1) d]=256 \\ & =3[2 a+5 d]=368[2 a+15 d]=256 \\ & =2 a+5 d=12 \ldots(1) \\ & 2 a+15 d=32 \ldots(2) \\ & \text { from }(1) \text { and }(2) \\ & -=-2 \\ & 10 d=20 \\ & d=2 \\ & P u t d=2 \text { in } \\ & 2 a+5(2)=12 \\ & 2 a=12-10 \\ & a=1 \\ & \left(\left[U \operatorname{sing} S_n=\frac{n}{2}[2 a+(n-1) d]\right]\right. \\ & \text { Putn }=1 \\ & =5[2+18] \\ & 5[20] \\ & S_{10}=100\end{aligned}$

Question:29

Find the sum of all the 11 terms of an AP whose middlemost term is 30.

Answer:

The middle term of an $\mathrm{AP}=\frac{n+1}{2}$
Here $\mathrm{n}=11$
Hence $\mathrm{a}_6=30$

$
\begin{aligned}
& a+5 d=30 \ldots(1)\left(U \operatorname{sing} a_{\mathrm{n}}=a+(n-1) d\right) \\
& S_n=\frac{n}{2}[2 a+(n-1) d] \text { putn }=11 \\
& =\frac{11}{2}[2 a+10 d] \\
& =\frac{n}{2}[2(a+5 d)] \\
& =11[30](U \operatorname{sing}(1)) \\
& S_{11}=330
\end{aligned}
$

Question:30

Find the sum of the last ten terms of the AP: 8, 10, 12, __________ , 126.

Answer:

The given AP is $8,10, \ldots, 126$
AP from last is $126,124, \ldots, 10,8$

$
a=126, \quad d=124-126=-2
$

The sum of 10 terms from the last:

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

Put $n=10$ :

$
\begin{gathered}
S_{10}=\frac{10}{2}[2(126)+(10-1)(-2)] \\
=5[252+9(-2)] \\
=5[252-18] \\
=5 \times 234 \\
S_{10}=1170
\end{gathered}
$

Question:31

Find the sum of the first seven numbers which are multiples of 2 as well as of 9.

Answer:

$\\\text{LCM of 2 and 9 is 18.\\ So the terms are 18, 36, } \ldots .., 126\\ \text{First term (a) }= 18\\$
$\text{Common difference (d)} = 36 - 18 = 18\\$
${{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right]$
$\\ Put\ n=7\\$
$\\ =\frac{7}{2}\left[ 36+108 \right] \\ =\frac{7}{2}\left[ 144 \right] \\$
$\\S_{7} = 7 \times 72 \\ S_{7} = 504\\$

Question:32

How many terms of the AP: -15, -13, -11, _______ are needed to make the sum -55? Explain the reason for the double answer.

Answer:

The given AP is $-15,-13, \ldots$
First term $a=-15$
Common difference $d=-13-(-15)=-13+15=2$
Let $n$ terms have a sum of -55 , then

$
\begin{gathered}
S_n=-55 \\
S_n=\frac{n}{2}[2 a+(n-1) d] \\
n[-30+2 n-2]=-55 \times 2 \\
-32 n+2 n^2+110=0 \\
2 n^2-32 n+110=0 \\
n^2-16 n+55=0 \\
n^2-5 n-11 n+55=0 \\
n(n-5)-11(n-5)=0 \\
(n-5)(n-11)=0 \\
n=5,11
\end{gathered}
$

There are two possible values for $n$ :
- When $n=5$, all terms are negative.
- When $n=11$, the AP will contain both positive and negative terms, leading to a sum of -55.

Question:33

The sum of the first n terms of an AP whose first term is 8 and the common difference is 20 is equal to the sum of the first 2n terms of another AP whose first term is 30 and the common difference is 8. Find n.

Answer:

The given first AP has:

$
a=8, d_1=20
$

The given second $A P$ has:

$
b=-30, d_2=8
$

It is given that $S_n=S_{2 n}$, so

$
\frac{n}{2}\left[2 a+(n-1) d_1\right]=\frac{2 n}{2}\left[2 b+(2 n-1) d_2\right]
$

Substituting values:

$
\begin{aligned}
\frac{n}{2}[2(8)+(n-1) 20] & =n[2(-30)+(2 n-1)(8)] \\
\frac{n}{2}[16+20 n-20] & =n[-60+16 n-8] \\
\frac{n}{2}[20 n-4] & =n[-68+16 n] \\
\frac{2(10 n-2)}{2} & =-68+16 n \\
10 n-16 n & =-68+2 \\
-6 n & =-66 \\
n=\frac{66}{6} & =11
\end{aligned}
$

Thus, $n=11$.

Question:34

Kanika was given her pocket money on Jan 1st, 2008. She puts Re 1 on Day 1, Rs 2 on Day 2, Rs 3 on Day 3, and continues 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 found that at the end of the month, she still had Rs 100 with her. How much was her pocket money for the month?

Answer:

According to the question, she deposits Rs. 1 on Day 1, Rs. 2 on Day 2, and so on. Thus, the sequence follows: $1,2,3, \ldots, 31$, where $a=1, d=1, a_n=31$, and $n=31$.

The sum of an arithmetic series is given by:

$
S_n=\frac{n}{2}\left(a+a_n\right)
$

Substituting the values:

$
S_{31}=\frac{31}{2}(1+31)=\frac{31}{2} \times 32=31 \times 16=496
$

Total money calculation:

$
\begin{gathered}
\text { Total money }=\text { Total deposited }+ \text { Spent }+ \text { Left } \\
=496+204+100=800
\end{gathered}
$

Thus, the total amount is Rs. 800.

Question:35

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?

Answer:

As per the question, she saves Rs. 32 in the first month, Rs. 36 in the second month, and so on.
Thus, the sequence follows: $32,36,40, \ldots$
Here, the first term $a=32$, and the common difference $d=36-32=4$.
Let her savings reach Rs. 2000 in $n$ months.
The sum of an arithmetic series is given by:

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

Substituting the values:

$
\begin{gathered}
2000=\frac{n}{2}[2 \times 32+(n-1) \times 4] \\
2000=\frac{n}{2}[64+4 n-4] \\
2000=\frac{n}{2}[60+4 n] \\
4000=n[60+4 n] \\
4 n^2+60 n-4000=0
\end{gathered}
$

Dividing by 4 :

$
n^2+15 n-1000=0
$

Solving by factorization:

$
\begin{gathered}
n^2+40 n-25 n-1000=0 \\
n(n+40)-25(n+40)=0 \\
(n+40)(n-25)=0
\end{gathered}
$

Since $n=-40$ is not possible, we take $n=25$.
Hence, it takes 25 months to save Rs. 2000.

Question:1

The sum of the first five terms of an AP and the sum of the first seven terms of the same AP is 167. If the sum of the first ten terms of this AP is 235, find the sum of its first twenty terms.

Answer:

$
S_n=\frac{n}{2}(2 a+(n-1) d)
$
Using this formula for $S_5$ :

$
\begin{gathered}
S_5=\frac{5}{2}(2 a+4 d)=167 \\
5(2 a+4 d)=334 \Longrightarrow 2 a+4 d=66
\end{gathered}
$
Using this formula for $S_7$ :

$
\begin{gathered}
S_7=\frac{7}{2}(2 a+6 d)=167 \\
7(2 a+6 d)=334 \Longrightarrow 2 a+6 d=48
\end{gathered}
$
Subtract equation (1) from equation (2):

$
\begin{gathered}
(2 a+6 d)-(2 a+4 d)=48-66 \\
2 d=-18 \Longrightarrow d=-9
\end{gathered}
$

Substitute $d=-9$ into equation (1):

$
\begin{gathered}
2 a+4(-9)=66 \\
2 a-36=66 \Longrightarrow 2 a=102 \Longrightarrow a=51
\end{gathered}
$
Now, to find the sum of the first 20 terms:

$
\begin{gathered}
S_{20}=\frac{20}{2}(2 a+19 d) \\
S_{20}=10(2(51)+19(-9)) \\
S_{20}=10(102-171)=10(-69)=-690
\end{gathered}
$
Thus, the sum of the first 20th terms is -690.

Question:2

i)Find the sum of those integers between 1 and 500 which are multiples of 2 as well as of 5.
ii)Find the sum of those integers from 1 to 500 which are multiples of 2 as well as of 5.
iii)Find the sum of those integers from 1 to 500 which are multiples of 2 or 5.

Answer:

i)

The numbers that are multiples of both 2 and 5 are the multiples of 10 :

$
10,20,30, \ldots, 490
$

Here, the first term $a=10$, the common difference $d=20-10=10$, and the last term $a_n=490$.
To find $n$, use the formula for the $n$th term of an arithmetic sequence:

$
\begin{gathered}
a+(n-1) d=490 \\
10+(n-1) \times 10=490 \\
(n-1) \times 10=480 \\
n-1=\frac{480}{10}=48 \\
n=48+1=49
\end{gathered}
$

Now, using the sum formula for an arithmetic series:

$
\begin{aligned}
S_n & =\frac{n}{2}\left[a+a_n\right] \\
S_{49} & =\frac{49}{2}[10+490] \\
& =\frac{49}{2} \times 500 \\
=49 & \times 250=12,250
\end{aligned}
$

Thus, the sum of the series is 12,250 .

ii)

The numbers that are multiples of both 2 and 5 are multiples of 10:

$
10,20,30, \ldots, 500
$

Here, the first term $a=10$ and the common difference $d=20-10=10$.
The last term is $a_n=500$.
Using the formula for the $n$th term of an arithmetic sequence:

$
\begin{gathered}
a_n=a+(n-1) d \\
10+(n-1) \times 10=500 \\
(n-1) \times 10=490 \\
n-1=\frac{490}{10}=49 \\
n=49+1=50
\end{gathered}
$

Now, using the sum formula for an arithmetic series:

$
S_n=\frac{n}{2}\left[a+a_n\right]
$

Substituting $n=50$ :

$
\begin{aligned}
& S_{50}=\frac{50}{2}[10+500] \\
& =25 \times 510=12,750
\end{aligned}
$

Thus, the sum of the series is 12,750 .

iii)

The numbers that are multiples of 2 :
$2,4,6,8, \ldots, 500$
Here, the first term $a_1=2$, common difference $d_1=4-2=2$, and last term $a_{n_1}=500$.
To find $n_1$ :

$
n_1=\frac{500}{2}=250
$

Using the sum formula for an arithmetic series:

$
\begin{gathered}
S_{n_1}=\frac{n_1}{2}\left[a_1+a_{n_1}\right] \\
S_{n_1}=\frac{250}{2}[2+500]=125 \times 502=62,750
\end{gathered}
$

The numbers that are multiples of 5 :

$
5,10,15, \ldots, 500
$

Here, the first term $a_2=5$, common difference $d_2=10-5=5$, and last term $a_{n_2}=500$. To find $n_2$ :

$
n_2=\frac{500}{5}=100
$

Using the sum formula:

$
\begin{gathered}
S_{n_2}=\frac{n_2}{2}\left[a_2+a_{n_2}\right] \\
S_{n_2}=\frac{100}{2}[5+500]=50 \times 505=25,250
\end{gathered}
$

The numbers that are multiples of 10 (to be subtracted as they are counted in both multiples of 2 and 5):

$
10,20,30, \ldots, 500
$

Here, the first term $a_3=10$, common difference $d_3=20-10=10$, and last term $a_{n_3}=500$. To find $n_3$ :

$
n_3=\frac{500}{10}=50
$

Using the sum formula:

$
\begin{gathered}
S_{n_3}=\frac{n_3}{2}\left[a_3+a_{n_3}\right] \\
S_{n_3}=\frac{50}{2}[10+500]=25 \times 510=12,750
\end{gathered}
$

Now, applying the principle of inclusion-exclusion:
Sum of numbers that are multiples of 2 or $5=S_{n_1}+S_{n_2}-S_{n_3}$

$
\begin{aligned}
& =62,750+25,250-12,750 \\
& =88,000-12,750=75,250
\end{aligned}
$

Thus, the sum of integers from 1 to 500 that are multiples of 2 or 5 is 75,250 .

Question:3

The eighth term of an AP is half its second term, and the eleventh term exceeds
one-third of its fourth term by 1. Find the 15th term.

Answer:

Let the first term be $a$ and the common difference be $d$.
Given conditions:
1. $T_8=\frac{1}{2} T_2$ implies:

$
a+7 d=\frac{1}{2}(a+d) \Longrightarrow 2 a+14 d=a+d \Longrightarrow a=-13 d
$

2. $T_{11}=\frac{1}{3} T_4+1$ implies:

$
a+10 d=\frac{1}{3}(a+3 d)+1 \Longrightarrow 3 a+30 d=a+3 d+3 \Longrightarrow 2 a=-27 d+3
$
Substitute $a=-13 d$ into $2 a=-27 d+3$ :

$
2(-13 d)=-27 d+3 \Longrightarrow-26 d=-27 d+3 \Longrightarrow d=3
$
Substitute $d=3$ into $a=-13 d$ :

$
a=-39
$
Find the 15th term:

$
T_{15}=a+14 d=-39+14(3)=-39+42=3
$
Thus, the 15th term is 3.

Question:4

An AP consists of 37 terms. The sum of the three middlemost terms is 225, and the sum of the last three is 429. Find the AP.

Answer:

Total terms $=37$, middle term $=$

$
\frac{37+1}{2}=19
$
Three middle most terms, 18th, 19th and 20th, given

$
a_{18}+a_{19}+a_{20}=225
$

$
a+17 d+a+18 d+a+19 d=225\left(\text { Using } \mathrm{a}_{\mathrm{n}}=a+(n-1) d\right)
$

$
3 a+54 d=225
$

Dividing by 3, we get

$
\begin{aligned}
& a+18 d=75 \\
& a=75-18 d \ldots(1)
\end{aligned}
$

According to the question

$
\begin{aligned}
& a_{35}+a_{36}+a_{37}=429\left(\mathrm{Using} \mathrm{a}_{\mathrm{n}}=a+(n-1) d\right) \\
& a+34 d+a+35 d+a+36 d=429 \\
& 3 a+105 d=429
\end{aligned}
$

Dividing by three, we get

$
a+35 d=143
$ Putting value from equation (1), we get

$
\begin{aligned}
& 75-18 d+35 d=143 \\
& 17 d=143-75 \\
& d=68 \\
& d=\frac{68}{17} \quad d=4
\end{aligned}
$ Puttingt$ $d=4$ in (1), we get

$
\begin{aligned}
& a=75-72=3 \\
& a+d=3+4=7 \\
& a+2 d=3+2(4)=11
\end{aligned}
$
Required AP is $3,7,11, \ldots \ldots$

Question:5

i)Find the sum of the integers between 100 and 200 that are divisible by 9
ii)Find the sum of the integers between 100 and 200 that are not divisible by 9

Answer:

i)

Numbers between 100 and 200 which is divisible by 9 are $108,117,126, \ldots 198$.
Here $a=108, d=117-108=9$

$
\begin{aligned}
& a_{\mathrm{n}}=198 \\
& a+(n-1) \times d=198 \\
& 108+(n-1) \times 9=198 \\
& (n-1) \times 9=198-108 \\
& (n-1)=\frac{90}{9} \\
& n=10+1=11 \\
& \mathrm{~S}_{\mathrm{n}}=\frac{n}{2}[2 \mathrm{a}+(\mathrm{n}-1) \mathrm{d}]
\end{aligned}
$
Put $\mathrm{n}=11$

$
\begin{aligned}
\mathrm{S}_{11} & =\frac{11}{2}[2(108)+(11-1) 9] \\
& =\frac{11}{2}[216+90] \\
& =\frac{11}{2} \times 306 \\
& =11 \times 153 \\
\mathrm{~S}_{11} & =1683
\end{aligned}
$

ii)

Numbers between 100 and 200 are 101, 102, 103...., 199

$
\begin{aligned}
& \text { Here } \mathrm{a}=101, \mathrm{~d}=102 \cdot 101=1 \\
& a_{\mathrm{n}}=199 \\
& a+(n-1) \cdot d=199\left(U \operatorname{sing} a_{\mathrm{n}}=a+(n-1) d\right) \\
& 101+(n-1) 1=199 \\
& (n-1)=199-101 \\
& n=98+1 \\
& n=99 \\
& S_{99}=\frac{99}{2}[2 \times 101+(99-1)(1)] \\
& =\frac{99}{2}[202+98]=\frac{99}{2}[300] \\
& =99[150] \\
& =14850
\end{aligned}
$

Numbers between 100 and 200 which are divisible by 9 are 108, 117, 126, ... 198

$
\begin{aligned}
& \text { Here } b=108 \\
& d_1=117-108=9 \\
& b_{\mathrm{N}}=198 \\
& b+(N-1) \times d_1=198 \\
& 108+(N-1) \times 9=198 \\
& (N-1) \times 9=198-108 \\
& (N-1)=\frac{90}{9}
\end{aligned}
$

$
\begin{aligned}
& N=10+1=11 \\
& \mathrm{~S}_{\mathrm{N}}=\frac{\mathrm{N}}{2}\left[2 \mathrm{~b}+(\mathrm{N}-1) \mathrm{d}_1\right]
\end{aligned}
$

Put N $=11$

$
\begin{aligned}
\\
\begin{aligned}
\mathrm{S}_{11} & =\frac{11}{2}[2(108)+(11-1) 9] \\
& =\frac{11}{2}[216+90] \\
& =\frac{11}{2} \times 306 \\
& =11 \times 153 \\
\mathrm{~S}_{11} & =1683
\end{aligned}
\end{aligned}
$

The sum of the integers between 100 and 200 that are not divisible by 9
$=$ Sum of all numbers - the sum of numbers which is divisible by 9

$
\begin{aligned}
& =14850-1683 \\
& =13167
\end{aligned}
$

Question:6

The ratio of the 11th term to the 18th term of an AP is 2/3. Find the ratio of the 5th term to the 21st term, and also the ratio of the sum of the first five terms to the sum of the first 21 terms.

Answer:

According to the question

$
\begin{aligned}
& \frac{a_{11}}{a_{18}}=\frac{2}{3} \Rightarrow \frac{a+10 d}{a+17 d}=\frac{2}{3}\left(U \operatorname{sing} a_{\mathrm{n}}=a+(n-1) \cdot d\right) \\
& 3 a+30 d=2 a+34 d \\
& 3 a-2 a+30 d-34 d=0 \\
& a-4 d=0 a=4 d \ldots \text { (1) }
\end{aligned}
$

Ratio of $5^{\text {th }}$ and 21 th term is

$
\frac{a_5}{a_{21}}=\frac{a+4 d}{a+20 d} \ldots(2)
$

Put $\mathrm{a}=4 \mathrm{~d}$ in (2) we get $=\frac{4 d+4 d}{4 d+20 d}=\frac{8 d}{24 d}$

$
\frac{a_5}{a_{21}}=\frac{1}{3}
$

Ratio of $S_5$ to $S_{21}$ is $\frac{S_5}{S_{21}}=\frac{\frac{5}{2}[2 a+4 d]}{2[2 a+20 d]}$

$
\begin{aligned}
{\left[\because S_n\right.} & \left.=\frac{n}{2}[2 a+(n-1) d]\right] \\
\frac{S_5}{S_{21}} & =\frac{5[2 a+4 d]}{21[2 a+200]}
\end{aligned}
$

Put $a=4 d$ in equation (3) we get

$
\begin{gathered}
\frac{S_5}{S_{21}}=\frac{5[4 d d)+4 d]}{21[2(4 d)+20 d]} \\
=\frac{5[[d d+4 d]}{21[8 d+20 d]} \\
=\frac{5[12 d]}{21[28 d]} \\
=\frac{60 d}{588 d} \\
\frac{S_5}{S_{12}}=\frac{5}{49}
\end{gathered}
$

Question:7

Show that the sum of an AP whose first term is a, the second term b and the last term c, is equal to
$\frac{(a+c)(b+c-2 a)}{2(b-a)}$

Answer:

Here

$
\begin{aligned}
& a_1=a, a_2=b \\
& d=a_2-a_1 \\
& =b-a \\
& a_{\mathrm{n}}=c \quad \because a_{\mathrm{n}}=a+(n-1) d \\
& a_1+(n-1) \cdot d=c \quad \text { Put } \mathrm{a}_1=a, d=b-a \\
& (n-1)(b-a)=c-a \\
& \mathrm{n}-1=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{~b}-\mathrm{a}} \\
& \mathrm{n}=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{~b}-\mathrm{a}+1} \\
& \mathrm{n}=\frac{\mathrm{c}-\mathrm{a}+\mathrm{b}-\mathrm{a}}{\mathrm{~b}-\mathrm{a}} \\
& \mathrm{n}=\frac{\mathrm{c}+\mathrm{b}-2 \mathrm{a}}{\mathrm{~b}-\mathrm{a}} \\
& \mathrm{~S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left[2 \mathrm{a}_1+(\mathrm{n}-1) \mathrm{d}\right] \\
& =\frac{\mathrm{c}+\mathrm{b}-2 \mathrm{a}}{2(\mathrm{~b}-\mathrm{a})}[\mathrm{a}+\mathrm{a}+(\mathrm{n}-1) \mathrm{d}] \\
& =\frac{(\mathrm{c}+\mathrm{b}-2 \mathrm{a})}{2(\mathrm{~b}-\mathrm{a})}\left[\mathrm{a}+\mathrm{a}_{\mathrm{n}}\right] \quad\left[\because \mathrm{a}+(\mathrm{n}-1) \mathrm{d}=\mathrm{a}_{\mathrm{n}}\right] \\
& \frac{(\mathrm{a}+\mathrm{c})(\mathrm{b}+\mathrm{c}-2 \mathrm{a})}{2(\mathrm{~b}-\mathrm{a})} \quad\left[\because \mathrm{a}_{\mathrm{n}}=\mathrm{c}\right]
\end{aligned}
$

Hence proved.

Question:8

Solve the equation- 4 + (-1) + 2 +...+ x = 437

Answer:

Given series:

$
-4+(-1)+2+\ldots+x=437
$
Here $\mathrm{a}=-4$

$
\begin{aligned}
& d=-1-(-4) \\
& -1+4=+3 \\
& S_{\mathrm{n}}=437 \text { (given) }
\end{aligned}
$
We know that $S_n=\frac{n}{2}[2 a+(n-1) d]$

$
\begin{aligned}
& 437=\frac{n}{2}[2(-4)+(n-1) 3] \\
& 874=n[-8+3 n-3] \\
& 874=3 n^2-11 n \\
& 3 n^2-11 n-874=0 \\
& 3 n^2-57 n+46 n-874=0 \\
& 3 n(n-19)+46(n-19)=0 \\
& \begin{array}{l}
(n-19)(3 n+46)=0 \\
n-19=0 \quad 2 n+16=0 \\
n=19 \quad n=-46 / 3
\end{array}
\end{aligned}
$
Hence $n=19$

$
\begin{aligned}
& a_{\mathrm{n}}=x \\
& a+(n-1) d=x\left(\text { using } \mathrm{a}_{\mathrm{n}}=a+(n-1) d\right) \\
& -4+(19-1) 3=x \\
& -4+(18) 3=x \\
& -4+54=x \\
& x=50
\end{aligned}
$

Question:9

Jaspal Singh repays his total loan of Rs 118000 by paying every month, starting with the first instalment of Rs 1000. If he increases the instalment by Rs 100 every month, what amount will be paid by him in the 30th instalment? What amount of loan does he still have to pay after the 30th instalment?

Answer:1

Total loan = 118000
First installment (a) = 1000
Installment increased = 100
Hence d = 100
$30_{th} \text{installment} = a_{30} = a + 29d \\$
= 1000 + 29(100)
= 1000 + 2900
= 3900
Amount of loan paid after 30 instalments
= Total loan - 30th installments is total loan

$\begin{gathered}118000-\frac{30}{2}[2 \times 1000+(30-1) \times 100] \\ 118000-15[2000+2900] \\ 118000-15[4900] \\ 118000-73500=44500\end{gathered}$

Question:10

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 2 m. The flags are stored at the position of the middlemost 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 to collect her books? What is the maximum distance she travelled carrying a flag?

Answer:

Let her fix 13 flags on one side and 13 on the other side of the middle flag. The distance covered for fixing the first flag and returning is 2m + 2m = 4.

The distance covered for fixing the second flag and returning is 4m + 4m = 8m.

The distance covered for fixing the third flag and returning is 6m + 6m = 12m
Similarly, for thirteen flags is
4m + 8m + 12m + …….
Here, a = 4
d = 8 - 4 = 4
n = 13
$\begin{gathered}S_n=\frac{n}{2}[2 a+(n-1) d] \\ S_{13}=\frac{13}{2}[2(4)+(13-1) 4] \\ =\frac{13}{2}[56] \\ =364\end{gathered}$
Total distance covered = 364 × 2 = 728 m (for both side )
distance, only to carry the flag = 364.