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#### Question 1:

Choose the correct answer from the given four options:
In an AP, if d = –4, n = 7, an = 4, then a is
(A) 6
(B) 7
(C) 20
(D) 28

We know that
${a}_{n}=a+\left(n-1\right)d$
Given d = –4, n = 7, an = 4
$⇒4=a+\left(7-1\right)×\left(-4\right)\phantom{\rule{0ex}{0ex}}⇒4=a-24\phantom{\rule{0ex}{0ex}}⇒a=28$
Hence, the correct answer is option (D).

#### Question 2:

Choose the correct answer from the given four options:
In an AP, if a = 3.5, d = 0, n = 101, then an will be
(A) 0
(B) 3.5
(C) 103.5
(D) 104.5

We know that
${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}$
Given a = 3.5, = 0, = 101
$⇒{a}_{n}=3.5+\left(101-1\right)×0\phantom{\rule{0ex}{0ex}}⇒{a}_{n}=3.5+0\phantom{\rule{0ex}{0ex}}⇒{a}_{n}=3.5$
Hence, the correct answer is option (B).

#### Question 3:

Choose the correct answer from the given four options:
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

Given list of numbers – 10, – 6, – 2, 2,...
Let
We have
$\phantom{\rule{0ex}{0ex}}{a}_{2}-{a}_{1}=-6-\left(-10\right)=-6+10=4\phantom{\rule{0ex}{0ex}}{a}_{3}-{a}_{2}=-2-\left(-6\right)=-2+6=4\phantom{\rule{0ex}{0ex}}{a}_{4}-{a}_{3}=2-\left(-2\right)=2+2=4$
$⇒$common difference $d={a}_{2}-{a}_{1}={a}_{3}-{a}_{2}={a}_{4}-{a}_{3}=4$ is same
$⇒$It is an AP with common difference 4.
Hence, the correct answer is option (B).

#### Question 4:

Choose the correct answer from the given four options:
The 11th term of the AP: , ...is
(A) –20
(B) 20
(C) –30
(D) 30

${n}^{th}$ term of an AP is given by
${a}_{n}=a+\left(n-1\right)d$
Given AP:  , ...
where
$⇒{a}_{11}=-5+\left(11-1\right)×\frac{5}{2}\phantom{\rule{0ex}{0ex}}⇒{a}_{11}=-5+25\phantom{\rule{0ex}{0ex}}⇒{a}_{11}=20$
Hence, the correct answer is option (B).

#### Question 5:

Choose the correct answer from the given four options:
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

We know that
${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}$
where a = first term
d = common difference
n = no. of terms
an = nth term
Terms of an AP are given by
Given
$⇒$First term $a=-2$
$⇒$Second term ${a}_{2}=a+d=-2-2=-4$
$⇒$Third term ${a}_{3}=a+2d=-2+2\left(-2\right)=-2-4=-6$
$⇒$Fourth term ${a}_{4}=a+3d=-2+3\left(-2\right)=-2-6=-8$
$⇒$Four terms of an AP with given a and d are .
Hence, the correct answer is option (C).

#### Question 6:

Choose the correct answer from the given four options:
The 21st term of the AP whose first two terms are –3 and 4 is
(A) 17
(B) 137
(C) 143
(D) –143

Given
common difference $d={a}_{2}-{a}_{1}=4-\left(-3\right)=7$
nth term of an AP is given by ${a}_{n}=a+\left(n-1\right)d$
21st term of an AP is
${a}_{21}=-3+\left(21-1\right)×7\phantom{\rule{0ex}{0ex}}{a}_{21}=-3+140=137$
$⇒$21st term of an AP is 137.
Hence, the correct answer is option (B).

#### Question 7:

Choose the correct answer from the given four options:
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

nth term of an AP is given by
${a}_{n}=a+\left(n-1\right)d$
Given

and

subtracting (i) from (ii)
$3d=12\phantom{\rule{0ex}{0ex}}d=4$
from (i)
$a=13-4\phantom{\rule{0ex}{0ex}}a=9$
Now, 7th term of an AP is given by
${a}_{7}=9+\left(7-1\right)4\phantom{\rule{0ex}{0ex}}{a}_{7}=33$
$⇒$7th term of an AP is 3
Hence, the correct answer is option (B).

#### Question 8:

Choose the correct answer from the given four options:
Which term of the AP: 21, 42, 63, 84,... is 210?
(A) 9th
(B) 10th
(C) 11th
(D) 12th

Given AP: 21, 42, 63, 84, ......,210
where first term
common difference $d=42-21=21$
nth term an = 210
Now, to find n
we know that ${a}_{n}=a+\left(n-1\right)d$
$⇒210=21+\left(n-1\right)×21\phantom{\rule{0ex}{0ex}}⇒210-21=\left(n-1\right)×21\phantom{\rule{0ex}{0ex}}⇒189=\left(n-1\right)×21\phantom{\rule{0ex}{0ex}}⇒9=\left(n-1\right)\phantom{\rule{0ex}{0ex}}⇒n=10$
$⇒$210 is the 10th term of the given AP.
Hence, the correct answer is option (B).

#### Question 9:

Choose the correct answer from the given four options:
If the common difference of an AP is 5, then what is a18a13?
(A) 5
(B) 20
(C) 25
(D) 30

nth term of an AP is given by  ${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}$
Now to find a18 – a13
we have

and

subtracting (ii) from (i)
a18 – a13
Hence, the correct answer is option (C).

#### Question 10:

Choose the correct answer from the given four options:
What is the common difference of an AP in which a18a14 = 32?
(A) 8
(B) –8
(C) –4
(D) 4

nth term of an AP is given by  ${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}$
Now,

and

Given  a18 – a14 = 32
$⇒\left(a+17d\right)-\left(a+13d\right)=32\phantom{\rule{0ex}{0ex}}⇒a+17d-a-13d=32\phantom{\rule{0ex}{0ex}}⇒4d=32\phantom{\rule{0ex}{0ex}}⇒d=8$
$⇒$common difference of an AP is 8.
Hence, the correct answer is option (A).

#### Question 11:

Choose the correct answer from the given four options:
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

General terms of an AP are
1st AP with first term $-1$ and common difference d  is
2nd AP with first term  $-8$ and common difference d  is
Now, nth term of an AP is given by ${a}_{n}=a+\left(n-1\right)d$
4th term of first AP

4th term of second AP

subtracting (ii) from (i)

$⇒$Difference between their 4th term is 7.
Hence, the correct answer is option (C).

#### Question 12:

Choose the correct answer from the given four options:
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

nth term of an AP is given by ${a}_{n}=a+\left(n-1\right)d$
Given
$7{a}_{7}=11{a}_{11}\phantom{\rule{0ex}{0ex}}⇒7\left[a+\left(7-1\right)d\right]=11\left[a+\left(11-1\right)d\right]\phantom{\rule{0ex}{0ex}}⇒7\left(a+6d\right)=11\left(a+10d\right)\phantom{\rule{0ex}{0ex}}⇒7a+42d=11a+110d\phantom{\rule{0ex}{0ex}}⇒4a+68d=0\phantom{\rule{0ex}{0ex}}⇒a+17d=0\phantom{\rule{0ex}{0ex}}⇒a+\left(18-1\right)d=0\phantom{\rule{0ex}{0ex}}⇒{a}_{18}=0$
$⇒{18}^{th}$ term of an AP is 0.
Hence, the correct answer is option (D).

#### Question 13:

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

We know that
nth term from the end of an AP $=l-\left(n-1\right)d$
where = last term
d = common difference
n = no. of terms
Given AP: –11, –8, –5, ..., 49
where
4th term from the end = $49-\left(4-1\right)×3=49-9=40$
$⇒$4th term from the end of an AP is 40.
Hence, the correct answer is option (B).

#### Question 14:

Choose the correct answer from the given four options:
The famous mathematician associated with finding the sum of the first 100 natural numbers is
(A) Pythagoras
(B) Newton
(C) Gauss
(D) Euclid

Gauss is the famous mathematician associated with finding the sum of the first 100 natural numbers.
Hence, the correct answer is option (C).

#### Question 15:

Choose the correct answer from the given four options:
If the first term of an AP is –5 and the common difference is 2, then the sum of the first 6 terms is
(A) 0
(B) 5
(C) 6
(D) 15

Given first term of an AP $a=$–5 and the common difference $d=2$
Sum of n terms of an AP  ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
Now, sum of first 6 terms is given by
${S}_{6}=\frac{6}{2}\left[2×\left(-5\right)+\left(6-1\right)×2\right]\phantom{\rule{0ex}{0ex}}{S}_{6}=3\left(-10+10\right)\phantom{\rule{0ex}{0ex}}{S}_{6}=3×0=0$
$⇒$the sum of the first 6 terms is 0.
Hence, the correct answer is option (A).

#### Question 16:

Choose the correct answer from the given four options:
The sum of first 16 terms of the AP: 10, 6, 2,... is
(A) –320
(B) 320
(C) –352
(D) –400

Sum of n terms of an AP is $\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
Given AP: 10, 6, 2,...
where
Now, to find sum of first 16 terms
${S}_{16}=\frac{16}{2}\left[2×10+\left(16-1\right)×\left(-4\right)\right]\phantom{\rule{0ex}{0ex}}{S}_{16}=8\left(20-60\right)\phantom{\rule{0ex}{0ex}}{S}_{16}=8×\left(-40\right)\phantom{\rule{0ex}{0ex}}{S}_{16}=-320$
$⇒$sum of first 16 terms of an AP is $-320$
Hence, the correct answer is option (A).

#### Question 17:

Choose the correct answer from the given four options:
In an AP if a = 1, an = 20 and Sn = 399, then n is
(A) 19
(B) 21
(C) 38
(D) 42

Given in an AP a = 1, a= 20 and Sn = 399
we have a= 20

and Sn = 399

Hence, the correct answer is option (C).

#### Question 18:

Choose the correct answer from the given four options:
The sum of first five multiples of 3 is
(A) 45
(B) 55
(C) 65
(D) 75

First five multiples of 3 are 3, 6, 9, 12, 15
Their sum $3+6+9+12+15=45$
$⇒$The sum of first five multiples of 3 is 45.
Hence, the correct answer is option (A).

#### Question 1:

Which of the following form an AP? Justify your answer.
(i) –1, –1, –1, –1, ...
(ii) 0, 2, 0, 2, ...
(iii) 1, 1, 2, 2, 3, 3,...
(iv) 11, 22, 33,...
(v)
(vi) 2, 22, 23, 24, ...
(vii)

(vii) We know that sequence is an AP when the common difference is same.
Given sequence:
$\sqrt{12}-\sqrt{3}=2\sqrt{3}-\sqrt{3}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\sqrt{27}-\sqrt{12}=3\sqrt{3}-2\sqrt{3}=\sqrt{3}\phantom{\rule{0ex}{0ex}}\sqrt{48}-\sqrt{27}=4\sqrt{3}-3\sqrt{3}=\sqrt{3}$
we can observe that common difference is same.
Hence, the given sequence is an AP.

#### Question 2:

Justify whether it is true to say that –1, , ... forms an AP as a2a1 = a3a2.

For a sequence to be an AP, common difference must be same.
Given sequence: –1, , ...
$\frac{-3}{2}-\left(-1\right)=\frac{-3}{2}+1=\frac{-1}{2}\phantom{\rule{0ex}{0ex}}-2-\left(\frac{-3}{2}\right)=-2+\frac{3}{2}=\frac{-1}{2}\phantom{\rule{0ex}{0ex}}\frac{5}{2}-\left(-2\right)=\frac{5}{2}+4=\frac{13}{2}$
we can see that common difference is not same i.e, a2 – a1 = a3 – a2 ≠ ${a}_{4}-{a}_{3}$
$⇒$Given sequence does not form an AP.

#### Question 3:

For the AP: –3, –7, –11, ..., can we find directly a30a20 without actually finding a30 and a20? Give reasons for your answer.

Yes, we can directly find a30 – a20  without actual finding a30 and a20.
nth term of an AP is given by ${a}_{n}=a+\left(n-1\right)d$
Given AP: –3, –7, –11, ...
where first term $a=-3$ and common difference $d=-7-\left(-3\right)=-7+3=-4$
We have
a30 – a20

Hence, we have found a30 – a20 = $-40$ without actual finding a30 and a20.

#### 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?

General terms of an AP are
where a = first term and d = common difference
1st AP with first term 2 is
2nd AP with first term 7 is
The difference between any two corresponding terms is same because common difference is same between every corresponding term i.e, 5 $\left[7-2=\left(7+d\right)-\left(2+d\right)=\left(7+2d\right)-\left(2+2d\right)=5\right]$
So, the corresponding difference remain same if we keep the common difference is same for two AP.

#### Question 5:

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

Given AP: 31, 28, 25, ...
To check 0 is the term of the given AP
We know that ${a}_{n}=a+\left(n-1\right)d$
where
$⇒0=31+\left(n-1\right)×\left(-3\right)\phantom{\rule{0ex}{0ex}}⇒3\left(n-1\right)=31\phantom{\rule{0ex}{0ex}}⇒3n-3=31\phantom{\rule{0ex}{0ex}}⇒3n=34\phantom{\rule{0ex}{0ex}}⇒n=\frac{34}{3}$
$⇒$n is not a natural number.
Hence, 0 is not the term of the AP: 31, 28, 25, ...

#### Question 6:

The taxi fare after each km, when the fare is Rs 15 for the first km and Rs 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.

No, because the total fare after each km is
Let
Now,

$\phantom{\rule{0ex}{0ex}}{a}_{2}-{a}_{1}=23-15=8\phantom{\rule{0ex}{0ex}}{a}_{3}-{a}_{2}=31-23=8\phantom{\rule{0ex}{0ex}}{a}_{4}-{a}_{3}=39-31=8$
Since, all the successive terms of the given list have same difference i.e., common difference = 8

Hence, the total fare after each km form an AP.

#### Question 7:

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

(iv) Let Number of bacteria in food items are n.
Now, number of bacteria after every second when they double
$2n-n=n\phantom{\rule{0ex}{0ex}}4n-2n=2n\phantom{\rule{0ex}{0ex}}8n-4n=4n$
common difference is not same
Hence, it does not form an AP.

#### Question 8:

Justify whether it is true to say that the following are the nth terms of an AP.
(i) 2n – 3
(ii) 3n2+ 5
(iii) 1 + n + n2

(i) Yes, here an = 2– 3
Now, n = 1, a1 = 2 × (1) – 3 = ​– 1
n = 2, a2 =  2 × (2) – 3 = ​1,
n = 3, a3 =  2 × (3) – 3 = ​3,
n = 4, a4 = 2 × (4) – 3 = ​5,
Thus, the number series will be ​– 1, 1, 3, 5, ...
Here, a2 – a1 = 1 –  (– 1) = 2
a3 – a2 = 3 –  (1) = 2
a4​ – a3 = 3 –  (1) = 2
Since, common difference is same
Hence, 2– 3 is the nth terms of an AP.

(ii)  No, here an = 3n2 + 5
Now, n = 1, a1 = 3 × (1)2 + 5 = ​8
n = 2, a2 = 3 × (2)2 + 5 = ​17
n = 3, a3 = 3 × (3)2 + 5 = ​32
n = 4, a4 = 3 × (4)2 + 5 = ​69
Thus, the number series will be ​8, 17, 32, 69, ...
Here, a2 – a1 = 17 –  8 = 9
a3 – a2 = 32 –  (17) = 15
a4​ – a3 = 69 –  (32) = 37
Since, common difference is not same.
Hence, 3n2+ 5 is not the nth terms of an AP.

(iii) No, here an =  1 + n2
Now, n = 1, a1 = 1 + 1 + (1)2 = 3
n = 2, a2 = 1+ (2) + (2)2 = 7
n = 3, a3 = 1+ 3 + (3)2 = 13
n = 4, a4 = 1+ 4 + (4)2 = 21
Thus, the number series will be ​3, 7, 13, 21, ...
Here, a2 – a1 = 1 –  (– 1) = 2
a3 – a2 = 3 –  (1) = 2
a4​ – a3 = 3 –  (1) = 2
Since, common difference is not same.
Hence, 1 + n2 is not the nth terms of an AP.

#### Question 1:

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

 Column A Column B (A1) 2, – 2, – 6, – 10, ... (B1) $\frac{2}{3}$ (A2) a = –18, n = 10, an = 0 (B2) –5 (A3) a = 0, a10 = 6 (B3) 4 (A4) a2 = 13, a4 = 3 (B4) –4 (B5) 2 (B6) $\frac{1}{2}$ (B7) 5

(A1) Given AP: 2, – 2, – 6, – 10, ...
∵ common difference $-2-2=-4$ i.e, (B4)

(A2) Given a = –18, n = 10, a= 0
we know that ${a}_{n}=a+\left(n-1\right)d$
$⇒0=-18+\left(10-1\right)d\phantom{\rule{0ex}{0ex}}⇒18=9d\phantom{\rule{0ex}{0ex}}⇒d=2$
∵ common difference is 2 i.e, (B5)

(A3) Given  a = 0, a10 = 6
we know that ${a}_{n}=a+\left(n-1\right)d$
${a}_{10}=6\phantom{\rule{0ex}{0ex}}⇒a+\left(10-1\right)d=6\phantom{\rule{0ex}{0ex}}⇒0+9d=6\phantom{\rule{0ex}{0ex}}⇒d=\frac{2}{3}$
∵ common difference is $\frac{2}{3}$ i.e, (B1)

(A4) we know that ${a}_{n}=a+\left(n-1\right)d$
Given a= 13
...(i)
and a4 = 3
$a+3d=3$  ...(ii)
subtracting (i) from (ii)
$2d=-10\phantom{\rule{0ex}{0ex}}⇒d=-5$
∵ common difference is $-5$ i.e, (B2).

#### Question 2:

Verify that each of the following is an AP, and then write its next three terms.

(i)

(ii)

(iii)

(iv) a + b, (a + 1) + b, (a + 1) + (b + 1), ...

(v) a, 2a + 1, 3a + 2, 4a + 3,...

(v) Given sequence a, 2a + 1, 3a + 2, 4+ 3,... is an AP if common difference is same.
$\left(2a+1\right)-a=a+1\phantom{\rule{0ex}{0ex}}\left(3a+2\right)-\left(2a+1\right)=3a+2-2a-1=a+1\phantom{\rule{0ex}{0ex}}\left(4a+3\right)-\left(3a+2\right)=4a+3-3a-2=a+1$
Hence, the given sequence is an AP as the common difference is same i.e, $a+1$

General terms of an AP are
where a = first term and d = common difference
Next three terms i.e, fifth, sixth and seventh terms are

#### Question 3:

Write the first three terms of the APs when a and d are as given below:
(i)

(ii) a = –5, d = –3

(iii)

General terms of an AP are
where a = first term and d = common difference

(i)Given
First three terms of the APs are

(ii) Given = –5, d = –3
First three terms of the APs are

(iii) Given
First three terms of the APs are

#### Question 4:

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

Given a, 7, b, 23,  are in AP.
$⇒$common difference will be same. i.e,

Hence, the values of

#### Question 5:

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

nth term of an AP is given by ${a}_{n}=a+\left(n-1\right)d$
Given ${a}_{11}=19$

and
${a}_{13}-{a}_{8}=20\phantom{\rule{0ex}{0ex}}⇒a+\left(13-1\right)d-\left[a+\left(8-1\right)d\right]=20\phantom{\rule{0ex}{0ex}}⇒a+12d-a-7d=20\phantom{\rule{0ex}{0ex}}⇒5d=20\phantom{\rule{0ex}{0ex}}⇒d=4$
from (i)
$a=19-10×4=19-40=-21$
General terms of an AP are
Hence, an AP with  is

#### Question 6:

The 26th, 11th and the last term of an AP are 0, 3 and $-\frac{1}{5}$, respectively. Find the common difference and the number of terms.

nth term of an AP is given by ${a}_{n}=a+\left(n-1\right)d$
Given:

and

Now, $a=-25d=-25×\frac{-1}{5}=5$
last term of an AP = $\frac{-1}{5}$
$⇒a+\left(n-1\right)d=\frac{-1}{5}\phantom{\rule{0ex}{0ex}}⇒5+\left(n-1\right)×\left(\frac{-1}{5}\right)=\frac{-1}{5}\phantom{\rule{0ex}{0ex}}⇒-\left(\frac{n-1}{5}\right)=\frac{-1}{5}-5\phantom{\rule{0ex}{0ex}}⇒\frac{n-1}{5}=\frac{26}{5}\phantom{\rule{0ex}{0ex}}⇒n-1=26\phantom{\rule{0ex}{0ex}}⇒n=27$
Hence, the common difference is $\frac{-1}{5}$ and number of terms are 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.

nth term of an AP is given by ${a}_{n}=a+\left(n-1\right)d$
Given

and

#### Question 8:

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

Given: first term a = 12 and
We know that
nth term of an AP is ${a}_{n}=a+\left(n-1\right)d$
Now,
$⇒\left[a+\left(7-1\right)d\right]=\left[a+\left(11-1\right)d\right]-24\phantom{\rule{0ex}{0ex}}⇒a+6d=a+10d-24\phantom{\rule{0ex}{0ex}}⇒4d=24\phantom{\rule{0ex}{0ex}}⇒d=6$

term of an AP is 126.

#### Question 9:

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

Proof: ${n}^{th}$ term of an AP is ${a}_{n}=a+\left(n-1\right)d$
We have

Now,

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.

Given AP: 7, 10, 13,....
To check 55 is a term of this AP or not
We have
nth term of an AP ${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}$

Yes, 55 is the term of the AP 7, 10, 13,...
and 55 is the 17th term of this AP.

#### Question 11:

Determine k so that k2+ 4k + 8, 2k2 + 3k + 6, 3k2 + 4k + 4 are three consecutive terms of an AP.

Given: k2+ 4k + 8, 2k2 + 3k + 6, 3k2 + 4k + 4 are three consecutive terms of an AP.
i.e, common difference will be same
$⇒\left(2{k}^{2}+3k+6\right)-\left({k}^{2}+4k+8\right)=\left(3{k}^{2}+4k+4\right)-\left(2{k}^{2}+3k+6\right)\phantom{\rule{0ex}{0ex}}⇒{k}^{2}-k-2={k}^{2}+k-2\phantom{\rule{0ex}{0ex}}⇒2k=0\phantom{\rule{0ex}{0ex}}⇒k=0$
Hence, the value of k is 0.

#### Question 12:

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

Let  be the three parts in AP.
According to the question
$\left(a-d\right)+a+\left(a+d\right)=207\phantom{\rule{0ex}{0ex}}⇒a-d+a+a+d=207\phantom{\rule{0ex}{0ex}}⇒3a=207\phantom{\rule{0ex}{0ex}}⇒a=69$
and product of the two smallest parts = 4623
$⇒a\left(a-d\right)=4623\phantom{\rule{0ex}{0ex}}⇒69\left(69-d\right)=4623\phantom{\rule{0ex}{0ex}}⇒4761-69d=4623\phantom{\rule{0ex}{0ex}}⇒69d=138\phantom{\rule{0ex}{0ex}}⇒d=2$
Hence, three parts 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.

Let the angles of a triangle are .
Given: greatest angle is twice the least

and we know that

Hence, the angles of a triangle are .

#### 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.

Given: nth terms of the two APs: 9, 7, 5, ... and 24, 21, 18,... are the same
nth terms of an AP ${a}_{n}=a+\left(n-1\right)d$
For AP: 9, 7, 5, ....
first term = 9 and common difference = $7-9=-2\phantom{\rule{0ex}{0ex}}$
For AP: 24, 21, 18,...
first term = 24 and common difference = $21-24=-3$
According to the question,
$⇒\left[9+\left(n-1\right)×\left(-2\right)\right]=\left[24+\left(n-1\right)×\left(-3\right)\right]\phantom{\rule{0ex}{0ex}}⇒9-2n+2=24-3n+3\phantom{\rule{0ex}{0ex}}⇒11-2n=27-3n\phantom{\rule{0ex}{0ex}}⇒n=16$
Now, 16th term of an AP

Hence,

#### Question 15:

If 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.

nth term of an AP is given by ${a}_{n}=a+\left(n-1\right)d$
Given:

and

subtracting (i) from (ii)

Now, 10th term of an AP is

Hence, 10th term of an AP is $-1$.

#### Question 16:

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

We know that
nth term from the end of an AP = $l-\left(n-1\right)d\phantom{\rule{0ex}{0ex}}$
where = last term, d = common difference and n = number of terms
For AP: –2, –4, –6,..., –100.
l = –100, n = 12 and d = – 4 – (–2) = – 4 + 2 = –2
Now, 12th term from the end of an AP
$=-100-\left(12-1\right)×\left(-2\right)\phantom{\rule{0ex}{0ex}}=-100+22\phantom{\rule{0ex}{0ex}}=-78$
Hence, 12th term from end of an AP is $-78$.

#### Question 17:

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

We know that
nth term of an AP is given by ${a}_{n}=a+\left(n-1\right)d$
For AP: 53, 48, 43,...
a = 53, d = $48-53=-5$ and n = ?
To find first negative term
${a}_{n}<0\phantom{\rule{0ex}{0ex}}⇒a+\left(n-1\right)d<0\phantom{\rule{0ex}{0ex}}⇒53+\left(n-1\right)×\left(-5\right)<0\phantom{\rule{0ex}{0ex}}⇒53-5n+5<0\phantom{\rule{0ex}{0ex}}⇒58<5n\phantom{\rule{0ex}{0ex}}⇒5n>58\phantom{\rule{0ex}{0ex}}⇒n>\frac{58}{5}=11.6$
is a natural number .
$⇒n=12$
Hence, 12th term is the first negative term.

#### Question 18:

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

Given: Numbers between 10 and 300, which when divided by 4 leave a remainder 3 makes an AP
AP: 11, 15, 19, ....., 299
We know that
nth term of an AP is given by ${a}_{n}=a+\left(n-1\right)d$
For AP: 11, 15, 19, ....., 299
a = 11, d = $15-11=4$an = 299 and n = ?
$⇒299=11+\left(n-1\right)4\phantom{\rule{0ex}{0ex}}⇒288=4n-4\phantom{\rule{0ex}{0ex}}⇒4n=292\phantom{\rule{0ex}{0ex}}⇒n=73$
$⇒$73 numbers lie between 10 and 300, which when divided by 4 leave a remainder 3.

#### Question 19:

Find the sum of the two middle most terms of the AP: .

Given AP:
First we will find the number of terms
We know that
nth term of an AP is given by
${a}_{n}=a+\left(n-1\right)d$
For given AP,
$⇒\frac{13}{3}=\frac{-4}{3}+\left(n-1\right)×\frac{1}{3}\phantom{\rule{0ex}{0ex}}⇒\frac{13}{3}+\frac{4}{3}=\frac{\left(n-1\right)}{3}\phantom{\rule{0ex}{0ex}}⇒17=n-1\phantom{\rule{0ex}{0ex}}⇒n=18$

Now, the middle two terms are
To find: ${a}_{9}+{a}_{10}$
We have

Hence, the sum of the two middle most term is 3.

#### 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.

We know that
${S}_{n}=\frac{n}{2}\left(a+l\right)$
Given: first term a = –5, last term or l = 45, ${S}_{n}=120$
$⇒120=\frac{n}{2}\left(-5+45\right)\phantom{\rule{0ex}{0ex}}⇒240=40n\phantom{\rule{0ex}{0ex}}⇒n=6$
Now, ${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}$
$⇒45=-5+\left(6-1\right)d\phantom{\rule{0ex}{0ex}}⇒50=5d\phantom{\rule{0ex}{0ex}}⇒d=10$
Hence, the number of terms are 6 and the common difference is 10.

#### Question 21:

Find the sum:

(i) 1 + (–2) + (–5) + (–8) + ... + (–236)

(ii)

(iii) to 11 terms.

(i) 1 + (–2) + (–5) + (–8) + ... + (–236)
First term = 1, common difference d = –2 – 1 = –3, last term a= –236
First we will find the number of terms
We have
${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}-236=1+\left(n-1\right)×\left(-3\right)\phantom{\rule{0ex}{0ex}}-237=-3\left(n-1\right)\phantom{\rule{0ex}{0ex}}79=n-1\phantom{\rule{0ex}{0ex}}⇒n=80\phantom{\rule{0ex}{0ex}}$
Now, the sum of 80 terms
${S}_{n}=\frac{n}{2}\left(a+{a}_{n}\right)\phantom{\rule{0ex}{0ex}}{S}_{80}=\frac{80}{2}\left[1+\left(-236\right)\right]\phantom{\rule{0ex}{0ex}}{S}_{80}=40×\left(-235\right)\phantom{\rule{0ex}{0ex}}{S}_{80}=-9400\phantom{\rule{0ex}{0ex}}⇒{S}_{80}=-9400$

#### Question 22:

Which term of the AP: –2, –7, –12,... will be –77? Find the sum of this AP upto the term –77.

We know that
nth term of an AP ${a}_{n}=a+\left(n-1\right)d$
For Given AP: –2, –7, –12,....–77

$⇒-77=-2+\left(n-1\right)×\left(-5\right)\phantom{\rule{0ex}{0ex}}⇒-75=-5\left(n-1\right)\phantom{\rule{0ex}{0ex}}⇒15=n-1\phantom{\rule{0ex}{0ex}}⇒n=16$
∵ –77 is the 16th term of the given AP.
Now, to find the sum of these 16 terms
${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}⇒{S}_{16}=\frac{16}{2}\left[2×\left(-2\right)+\left(16-1\right)×\left(-5\right)\right]\phantom{\rule{0ex}{0ex}}⇒{S}_{16}=8\left(-4-75\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{16}=8×\left(-79\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{16}=-632$
Hence, the sum of 16 terms of the given AP is –632.

#### Question 23:

If an = 3 – 4n, show that a1, a2 ,a3 ,... form an AP. Also find S20.

Given: a= 3 – 4n
We have
${a}_{1}=3-4×1=-1\phantom{\rule{0ex}{0ex}}{a}_{2}=3-4×2=-5\phantom{\rule{0ex}{0ex}}{a}_{3}=3-4×3=-9\phantom{\rule{0ex}{0ex}}{a}_{4}=3-4×4=-13\phantom{\rule{0ex}{0ex}}.\phantom{\rule{0ex}{0ex}}.\phantom{\rule{0ex}{0ex}}.$
we can observe that
i.e, common difference$\left(d=-4\right)$ is same
Hence, a1, a2 ,a3 ,... form an AP.
Now to find the sum of 16 terms
${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}⇒{S}_{20}=\frac{20}{2}\left[2×\left(-1\right)+\left(20-1\right)×\left(-4\right)\right]\phantom{\rule{0ex}{0ex}}⇒{S}_{20}=10\left(-2-76\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{20}=10×\left(-78\right)\phantom{\rule{0ex}{0ex}}⇒{S}_{20}=-780$
Hence, the sum of 20 terms is –780.

#### Question 24:

In an AP, if Sn = n (4n + 1), find the AP.

Given: Sum of terms of an AP Sn = n (4n + 1)
We have
${S}_{1}=1\left(4×1+1\right)=5\phantom{\rule{0ex}{0ex}}{S}_{2}=2\left(4×2+1\right)=18\phantom{\rule{0ex}{0ex}}{S}_{3}=3\left(4×3+1\right)=39\phantom{\rule{0ex}{0ex}}{S}_{4}=4\left(4×4+1\right)=68\phantom{\rule{0ex}{0ex}}.$
We have
${a}_{1}={S}_{2}-{S}_{1}=18-5=13\phantom{\rule{0ex}{0ex}}{a}_{2}={S}_{3}-{S}_{2}=39-18=21\phantom{\rule{0ex}{0ex}}{a}_{3}={S}_{4}-{S}_{3}=68-39=29\phantom{\rule{0ex}{0ex}}.\phantom{\rule{0ex}{0ex}}.$]
Hence, an AP is 13, 21, 29, .....

#### Question 25:

In an AP, if Sn = 3n2 + 5n and ak = 164, find the value of k.

Given: Sum of terms of an AP Sn = 3n2 + 5n
We have

$⇒$First term a = 14, common difference d = 6
Given ak = 164
$⇒a+\left(k-1\right)d=164\phantom{\rule{0ex}{0ex}}⇒14+\left(k-1\right)6=164\phantom{\rule{0ex}{0ex}}⇒6\left(k-1\right)=150\phantom{\rule{0ex}{0ex}}⇒k-1=25\phantom{\rule{0ex}{0ex}}⇒k=26$
Hence, the value of k = 26.

#### Question 26:

If Sn denotes the sum of first n terms of an AP, prove that S12 = 3(S8 – S4)

To prove: S12 = 3(S8 – S4)
Proof:
We know that
Sum of  terms of an AP is
${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
Now,

and

Hence Proved.

#### Question 27:

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

Given:
We know that
nth term of an AP is
${a}_{n}=a+\left(n-1\right)d\phantom{\rule{0ex}{0ex}}$
Now,

subtracting (i) from (ii)

Now to find sum of  first 17 terms of an AP
${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{17}=\frac{17}{2}\left[2×\left(-6\right)+\left(17-1\right)×\left(-3\right)\right]\phantom{\rule{0ex}{0ex}}{S}_{17}=\frac{17}{2}\left(-12-48\right)\phantom{\rule{0ex}{0ex}}{S}_{17}=\frac{17}{2}×\left(-60\right)\phantom{\rule{0ex}{0ex}}{S}_{17}=-510$
Hence, sum of first 17 terms of an AP is $-530.$

#### Question 28:

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

We know that
Sum of first terms of an AP is
${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}$
Given:

and S16 = 256

Now sum of 10 terms
${S}_{10}=\frac{10}{2}\left[2×1+\left(10-1\right)×2\right]\phantom{\rule{0ex}{0ex}}{S}_{10}=5\left(2+18\right)\phantom{\rule{0ex}{0ex}}{S}_{10}=100$
Hence, the sum of 10 terms of an AP is 100.

#### Question 29:

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

To find: sum of all the 11 terms of an AP
Given: middle most term of 11 terms is

Now, to find the sum of 11 terms of an AP

Hence, the sum of 11 terms of an AP is 330.

#### Question 30:

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

Given AP: 8, 10, 12,---, 124, 126.
After reversing the AP
New AP is 126, 124, 122, ,---, 8
We have
First term a = 126, common difference d = $124-126=-2$number of terms n = 7
To find the sum of 10 terms
${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]\phantom{\rule{0ex}{0ex}}{S}_{10}=\frac{7}{2}\left[2×126+\left(10-1\right)×\left(-2\right)\right]\phantom{\rule{0ex}{0ex}}{S}_{10}=\frac{7}{2}\left(252-18\right)\phantom{\rule{0ex}{0ex}}{S}_{10}=\frac{7}{2}×234\phantom{\rule{0ex}{0ex}}{S}_{10}=819$
Hence, the sum of ten term of an AP is 819.

#### Question 31:

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

Seven numbers which are multiples of 2 as well as of 9 forms an AP i.e, 18, 36, 54, .....
First term a = 18, common difference d = 36 $-$18 = 18, number of terms n = 7
To find their sum
We have ${S}_{n}=\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
${S}_{7}=\frac{7}{2}\left[2×18+\left(7-1\right)×18\right]\phantom{\rule{0ex}{0ex}}{S}_{7}=\frac{7}{2}\left[36+108\right]\phantom{\rule{0ex}{0ex}}{S}_{7}=\frac{7}{2}×144\phantom{\rule{0ex}{0ex}}{S}_{7}=504$
Hence, sum of first seven numbers which are multiples of 2 as well as of 9 is 504.

#### Question 32:

How many terms of the AP: –15, –13, –11,--- are needed to make the sum –55?

Explain the reason for double answer.

We know that
Sum of terms of an AP is $\frac{n}{2}\left[2a+\left(n-1\right)d\right]$
Given AP: –15, –13, –11,---
where

#### 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 first 2n terms of another AP whose first term is – 30 and the common difference is 8. Find n.

Given, the first term of the first AP (a) = 8
and the common difference of the first AP (d) = 20
Let the number of terms in first AP be n.

Now, the first term of the second AP ( a' ) = $-$30
and the common difference of the second AP ( d' ) = 8

As per the condition given,
Sum of n terms of the first AP = Sum of the first 2n terms of the second AP.
From (1) and (2)
$⇒{S}_{n}={S}_{2n}\phantom{\rule{0ex}{0ex}}⇒n\left(10n-2\right)=n\left(16n-68\right)\phantom{\rule{0ex}{0ex}}⇒n\left[\left(16n-68\right)-\left(10n-2\right)\right]=0\phantom{\rule{0ex}{0ex}}⇒n\left[16n-68-10n+2\right]=0\phantom{\rule{0ex}{0ex}}⇒n\left(6n-66\right)=0\phantom{\rule{0ex}{0ex}}⇒n=11\phantom{\rule{0ex}{0ex}}$

Hence, the required value on 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 continued 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?

Let her pocket money be INR x

Now, she takes  INR 1 on day 1,  INR 2 on day 2,  INR 3 on day 3 and so on till the end of the month, from this money.
i.e.,                               1 + 2 + 3 + 4 + … + 31

which form an AP in which terms are 31 and first term (a) = 1, common difference (d) = 2 — 1 = 1 .

Sum of first 31 terms = S31

So kanika takes INR 496 amount at the end of the month.
She spent INR 204 of her pocket money and found INR 100 left with her at the end of the month.
As per the condition

Thus she got INR 800 as her pocket money for the month.

#### 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?

During the 1st month, she saves = INR 32

During the 2nd month, she saves = INR 36

During the 3rd month, she saves = INR 40

Say,  Yasmeen saves INR 2000 during the n months.

Here, we have arithmetic progression 32, 36, 40,…
First term (a) = 32, common difference (d) = 36 – 32 = 4
and she saves total money, i.e., Sn = INR 2000

We know, sum of first n terms of an AP is

Hence, she saves INR 2000 in 25 months.

#### 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.

Let the first term be a, common difference d and number of terms in an AP as n.
Then,

Multiplying (5) by 6 and subtracting (4), we get

Hence, the required sum of its first twenty terms is 970.

#### Question 2:

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

(i)
As the multiples of 2 as well as of 5 is the LCM of (2, 5) = 10

∴  Multiples of 2 as well as of 5 between 1 and 500 are 10, 20, 30, ..., 490.

Which is an AP with first term (a) = 10 and common difference (d) = 20 $-$ 10 = 10
and nth term  = Last term (l) = 490
∴ Sum of n terms between 1 and 500,

(ii) Multiples of 2 as well as of 5 from 1 to 500 are 10, 20, 30, ..., 500.
(a) = 10, (d) =10 and Last term (l) = 500

(iii) Multiples of 2 or 5 = List of multiples of 2 from 1 to 500 +  List of multiples of 5 from 1 to 500 $-$ List of multiples of 10 from 1 to 500

=(2, 4, 6, ...., 500) + (5, 10, 15, ..., 500) $-$ (10, 20, 30, ..., 500)                         ....(1)

All of these AP's form an AP in itself.
No. of terms in first AP =
500 = 2 + (n1 $-$ 1)2 ⇒ 498 = (n1 $-$ 1)2
(n1 $-$ 1) = 249 ⇒n1 = 250

No. of terms in second AP =
500 = 5 + (n2 $-$ 1)5 ⇒ 495 = (n2 $-$ 1)5
(n2 $-$ 1) = 99 ⇒n2 = 100

No. of terms in third AP =
500 = 10 + (n3 $-$ 1)10 ⇒ 490 = (n3 $-$ 1)10
(n3 $-$ 1) = 49 ⇒n3 = 50

From (1), sum of multiples of 2 or 5 from 1 to 500 is

#### 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.

Let the first term be a and common difference be d of an AP respectively.
Then,

Hence, the 15th term is 3.

#### Question 4:

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

The total term in the AP is an odd value. Thus,

Hence, the three middle terms are 18th, 19th and 20th
As per given situation,

Hence, the required AP is 3, 7, 11, 15, ....

#### Question 5:

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

(i) The numbers (integers) between 100 and 200 which is divisible by 9 are 108, 117, 126, … 198.
Consider the number of terms between 100 and 200 which is divisible by 9 be n.
Then,

Hence, the sum of the integers between 100 and 200 , divisible by 9 is 1683.

(ii) The sum of the integers between 100 and 200, not divisible by 9
= (sum of total numbers between 100 and 200) – (sum of total numbers between 100 and 200 , divisible by 9)

Sum of the integer between 100 and 200, not divisible by 9 is
= 14850 $-$ 1683
= 13167

Hence, the required sum is 13167

#### 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.

Let the first term = a and common difference = d . Then,

Hence, the ratio of the sum of the first 5 terms to the sum of first 21 terms is 5:49

#### 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 $\frac{\left(a+c\right)\left(b+c-2a\right)}{2\left(b-a\right)}$

The AP is a, b,..., c
Here, first term = a, common difference = b – a, and last term, = an = c.

Hence proved

#### Question 8:

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

Given AP is, – 4 + (–1) + 2 + … + x = 437                                                                                  …(i)
Here, – 4 + (–1) + 2 + …+ x forms an AP with first term = –4, common difference = 3,
an = l = x,

Here, x cannot be negative, i.e., x ≠ –53
also, for x = –53, n will be negative which is not possible

Hence, the required value of x is 50.

#### 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?

Jaspal Singh takes total loan = ₹ 118000 He repays his total loan by paying every month.

His first installment = ₹ 1000

Second installment = 1000 + 100 = ₹ 1100

Third installment = 1100 + 100 = ₹ 1200 and so on

The AP is 1000, 1100, 1200, ...., an .

Let its 30th instalment be n,

with first term (a) = 1000 and common difference (d) = 1100 – 1000 = 100 and nth term of an AP, Tn = a + (n – 1)d
For 30th instalment, T30 = 1000 + (30 – 1)100
= 100 + 29 × 100 = 1000 + 2900
T30 = 3900

So, ₹ 3900 will be paid by him in th 30th instalment.

He paid total amount upto 30 instalments in the following form 1000 + 1100 + 1200 + …………. + 3900
First term (a) = 1000 and last term (1) = 3900

Sum of 30 instalments, ${S}_{30}=\frac{30}{2}\left(n+l\right)$     [ ∵ sum of first n terms of an AP is, ${S}_{n}=\frac{n}{2}\left[a+l\right]$ where l = last term]
⇒ S30 = 15(1000 + 3900) = 15 × 4900 = ₹ 73500

The total amount he still has to pay after the 30 installment
= (Amount of loan) – (Sum of 30 installments)
= 11800073500 = INR 44500

Hence, after the 30th installment, he will still have to pay INR 44500.

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

The students of a school decided to beautify the school oh the annua] day by fixing colorful flags on the straight passage of the school.

Given, the number of flags = 27 and the distance between each flag = 2 m.

The flags are also stored in the position of the middle flag, which is the 14th flag, and Ruchi is in charge of installing the flags. Ruchi maintained her books, which were where the flags were kept, i.e., the 14th flag, and she could only carry one flag at a time.

She started with the middle flag, the 14th flag, and moved 13 flags to her left. Distance traveled to place the second flag and return to his original location = 2 + 2= 4 m.

Similarly, the distance traveled to place the third flag and return to his original position is = 4+ 4= 8 m. Distance traveled = 6 + 6= 12 m for setting the fourth flag and returning to his original place.

Distance traveled to place the fourteenth flag and return to his original position, = 26 + 26= 52 m

Continue in the same fashion into her right position, starting with the 14th flag. In that situation, the total distance traveled was 52 meters.

Ruchi also returned to his center position and collected her books after placing the final flag. This distance also included where the last flag was placed.

So, these distances form a series, 4 + 8 + 12 + 16 + …+ 52      [for left]
and    4 + 8 + 12 + 16 +… + 52     [for right]

Total distance covered by Ruchi for placing these flags

As a result, the required distance is 728 meters, which she covered while finishing this duty and returned to retrieve her books.

Ruchi's maximum distance traveled while carrying a flag is now equal to the distance she traveled while placing the 14th flag in her left position or the 27th flag in her right position.

= (2 + 2 + 2 + … + 13 times)
= 2 x 13 = 26m

Hence, the required maximum distance she traveled carrying a flag is 26 m.

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