# NCERT solution for class 10 maths arithmetic progressions ( Chapter 5) #### Solution for Exercise 5.1

1. In which of the following situations, does the list of numbers involved make an arithmetic progression, and why?
(i) The taxi fare after each km when the fare is Rs 15 for the first km and Rs 8 for each additional km.
(ii) The amount of air present in a cylinder when a vacuum pump removes $$\frac{1}{4}$$th of the air remaining in the cylinder at a time.
(iii) The cost of digging a well after every meter of digging, when it costs Rs 150 for the first meter and rises by Rs 50 for each subsequent meter.
(iv) The amount of money in the account every year, when Rs 10,000 is deposited at compound Interest at 8% per annum.

(i)Taxi fare for 1st km = Rs 15, Taxi fare after 2 km = 15 + 8 = Rs 23
Taxi fare after 3 km = 23 + 8 = Rs 31
Taxi fare after 4 km = 31 + 8 = Rs 39
Therefore, the sequence is 15, 23, 31, 39…
It is an arithmetic progression because difference between any two consecutive terms is equal which is 8.
(23 – 15 = 8, 31 – 23 = 8, 39 – 31 = 8, …)

(ii)Let amount of air initially present in a cylinder = V
Amount of air left after pumping out air by vacuum pump = $$V - {{V} \over {4}} = {{4V - V} \over {4}} = {{3V} \over {4}}$$
Amount of air left when vacuum pump again pumps out air
=$${{3V} \over {4}} - ( {{1} \over {4}} × {{3V} \over {4}} ) = {{3V} \over {4}} - {{3V} \over {16}} = {{12V - 3V} \over {16}} = {{9V} \over {16}}$$
So, the sequence we get is like $$V, {{3V} \over {4}}, {{9V} \over {16}}.....$$
Checking for difference between consecutive terms …
$${{3V} \over {4}} - V = {{-V} \over {4}}$$,$${{9V} \over {16}} - {{3V} \over {4}} = {{9V - 12V} \over {16}} = {{-3V} \over {16}}$$
Difference between consecutive terms is not equal.
Therefore, it is not an arithmetic progression.

(iii) Cost of digging 1 meter of well = Rs 150
Cost of digging 2 meters of well = 150 + 50 = Rs 200
Cost of digging 3 meters of well = 200 + 50 = Rs 250
Therefore, we get a sequence of the form 150, 200, 250…
It is an arithmetic progression because difference between any two consecutive terms is equal. (200 – 150 = 250 – 200 = 50…)
Here, difference between any two consecutive terms (common difference) is equal to 50.

(iv)Amount in bank after Ist year = $$1000 ( 1 + {{8} \over {100}})$$… (1)
Amount in bank after two years = $$1000 ( 1 + {{8} \over {100}})^2$$… (2)
Amount in bank after three years = $$1000 ( 1 + {{8} \over {100}})^3$$… (3)
Amount in bank after four years = $$1000 ( 1 + {{8} \over {100}})^4$$… (4)
It is not an arithmetic progression because $$(2) - (1) \ne (3) - (2)$$
(Difference between consecutive terms is not equal)
Therefore, it is not an Arithmetic Progression.

2. Write first four terms of the AP, when the first term a and common difference d are given as follows:
(i)a = 10, d = 10
(ii) a = -2, d = 0
(iii) a = 4, d = -3
(iv) a = -1, d = $${{1} \over {2}}$$
(v) a = -1.25, d = -0.25

(i) First term = a = 10, d = 10
Second term = a + d = 10 + 10 = 20
Third term = second term + d = 20 + 10 = 30
Fourth term = third term + d = 30 + 10 = 40
Therefore, first four terms are: 10, 20, 30, 40

(ii) First term = a = –2 , d = 0
Second term = a + d = –2 + 0 = –2
Third term = second term + d = –2 + 0 = –2
Fourth term = third term + d = –2 + 0 = –2
Therefore, first four terms are: –2, –2, –2, –2

(iii) First term = a = 4, d =–3
Second term = a + d = 4 – 3 = 1
Third term = second term + d = 1 – 3 = –2
Fourth term = third term + d = –2 – 3 = –5
Therefore, first four terms are: 4, 1, –2, –5

(iv) First term = a = –1, d = $${{1} \over {2}}$$
Second term = a + d = –1 + $${{1} \over {2}}$$ = $${{-1} \over {2}}$$
Third term = second term + d = $${{-1} \over {2}} + {{1} \over {2}}$$ = 0
Fourth term = third term + d = 0 + $${{1} \over {2}}$$ = $${{1} \over {2}}$$
Therefore, first four terms are: –1, $${{-1} \over {2}}$$, 0, $${{1} \over {2}}$$

(v) First term = a = –1.25, d = –0.25
Second term = a + d = –1.25 – 0.25 = –1.50
Third term = second term + d = –1.50 – 0.25 = –1.75
Fourth term = third term + d = –1.75 – 0.25 = –2.00
Therefore, first four terms are: –1.25, –1.50, –1.75, –2.00

3. For the following APs, write the first term and the common difference.
(i) 3, 1, –1, –3 …
(ii) –5, –1, 3, 7…
(iii) $${{1}\over{3}},{{5}\over{3}},{{9}\over{3}},{{13}\over{3}}$$
(iv) 0.6, 1.7, 2.8, 3.9 …

(i) 3, 1, –1, –3…
First term = a = 3,
Common difference (d) = Second term – first term = Third term – second term and so on
Therefore, Common difference (d) = 1 – 3 = –2

(ii) –5, –1, 3, 7…
First term = a = –5
Common difference (d) = Second term – First term
= Third term – Second term and so on
Therefore, Common difference (d) = –1 – (–5) = –1 + 5 = 4

(iii)$${{1}\over{3}},{{5}\over{3}},{{9}\over{3}},{{13}\over{3}}$$
First term = a = $${{1}\over{3}}$$
Common difference (d) = Second term – First term
= Third term – Second term and so on
Therefore, Common difference (d) = $${{5}\over{3}} - {{1}\over{3}} = {{4}\over{3}}$$

(iv) 0.6, 1.7, 2.8, 3.9…
First term = a = 0.6
Common difference (d) = Second term – First term
= Third term – Second term and so on
Therefore, Common difference (d) = 1.7 - 0.6 = 1.1

4. Which of the following are APs? If they form an AP, find the common difference d and write three more terms.
(i) $$2, 4, 8, 16…$$
(ii) 2, $${{5}\over{2}}$$ , 3, $${{7}\over{2}}$$…
(iii) $$-1.2, -3.2, -5.2, -7.2…$$
(iv) $$-10, -6, -2, 2…$$
(v) $$3, 3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}$$…
(vi) $$0.2, 0.22, 0.222, 0.2222…$$
(vii) $$0, -4, -8, -12…$$
(viii) $${{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}$$…
(ix) $$1, 3, 9, 27…$$
(x)$$a, 2a, 3a, 4a…$$
(xi) $$a , a^2 , a^3 , a^4$$…
(xii) $$\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}$$ …
(xiii) $$\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}$$…
(xiv) $$1^2, 3^2, 5^2, 7^2$$…
(xv) $$1^2, 5^2, 7^2, 73$$…

(i) 2, 4, 8, 16…
It is not an AP because difference between consecutive terms is not equal.
4 – 2 $$\ne$$ 8 - 4

(ii) 2, $${{5}\over{2}}$$ , 3, $${{7}\over{2}}$$…
It is an AP because difference between consecutive terms is equal.
=>$${{5}\over{2}} - 2 = 3 - {{5}\over{2}} = {{1}\over{2}}$$
Common difference (d) = $${{1}\over{2}}$$
Fifth term = $${{7}\over{2}} + {{1}\over{2}} = 4$$
Sixth term = 4 + $${{1}\over{2}} = {{9}\over{2}}$$
Seventh term = $${{9}\over{2}} + {{1}\over{2}} = 5$$
Therefore, next three terms are 4, $${{9}\over{2}}$$ and 5.

(iii)-1.2, -3.2, -5.2, -7.2…
It is an AP because difference between consecutive terms is equal.
-3.2 - (-1.2)
= -5.2 - (-3.2)
= -7.2 - (-5.2) = -2
Common difference (d) = -2
Fifth term = -7.2 – 2 = -9.2
Sixth term = -9.2 – 2 = -11.2
Seventh term = -11.2 – 2 = -13.2
Therefore, next three terms are -9.2, -11.2 and -13.2

(iv) -10, -6, -2, 2…
It is an AP because difference between consecutive terms is equal.
-6 - (-10) = -2 - (-6)
= 2 - (-2) = 4
Common difference (d) = 4
Fifth term = 2 + 4 = 6
Sixth term = 6 + 4 = 10
Seventh term = 10 + 4 = 14
Therefore, next three terms are 6, 10 and 14

(v) 3 ,$$3 + \sqrt{2}, 3 + 2\sqrt{2}, 3 + 3\sqrt{2}$$…
It is an AP because difference between consecutive terms is equal.
= $$3 + \sqrt{2} - 3 = 3 + 2\sqrt{2} - 3 + \sqrt{2} = \sqrt{2}$$
Common difference (d) = $$\sqrt{2}$$
Fifth term = $$3 + 3\sqrt{2} + \sqrt{2} = 3 + 4\sqrt{2}$$
Sixth term = $$3 + 4\sqrt{2} + \sqrt{2} = 3 + 5\sqrt{2}$$
Seventh term = $$3 + 5\sqrt{2} + \sqrt{2} = 3 + 6\sqrt{2}$$
Therefore, next three terms are $$3 + 4\sqrt{2}, 3 + 5\sqrt{2}. 3 + 6\sqrt{2}$$

(vi) 0.2, 0.22, 0.222, 0.2222…
It is not an AP because difference between consecutive terms is not equal.
0.22 - 0.2 $$\ne$$ 0.222 - 0.22

(vii) 0, -4, -8, -12…
It is an AP because difference between consecutive terms is equal.
-4 – 0 = -8 - (-4)
= -12 - (-8) = -4
Common difference (d) = -4
Fifth term = -12 – 4 =-16
Sixth term = -16 – 4 = -20
Seventh term = -20 – 4 = -24
Therefore, next three terms are -16, -20 and -24

(viii) $${{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}, {{-1}\over{2}}$$…
It is an AP because difference between consecutive terms is equal.
=$${{-1}\over{2}} - {{-1}\over{2}} = {{-1}\over{2}} - {{-1}\over{2}} = 0$$
Common difference (d) = 0
Fifth term = $${{-1}\over{2}} + 0 = {{-1}\over{2}}$$
Sixth term = $${{-1}\over{2}} + 0 = {{-1}\over{2}}$$
Seventh term =$${{-1}\over{2}} + 0 = {{-1}\over{2}}$$
Therefore, next three terms are $${{-1}\over{2}},{{-1}\over{2}},{{-1}\over{2}}$$

(ix) 1, 3, 9, 27…
It is not an AP because difference between consecutive terms is not equal.
3 – 1 $$\ne$$ 9 - 3

(x) a, 2a, 3a, 4a…
It is an AP because difference between consecutive terms is equal.
2a – a = 3a - 2a = 4a - 3a = a
Common difference (d) = a
Fifth term = 4a + a = 5a Sixth term = 5a + a = 6a
Seventh term = 6a + a = 7a
Therefore, next three terms are 5a, 6a and 7a

(xi) $$a , a^2 , a^3 , a^4$$…
It is not an AP because difference between consecutive terms is not equal
$$a^2 - a \ne a^3 - a^2$$

(xii) $$\sqrt{2}, \sqrt{8}, \sqrt{18}, \sqrt{32}$$…
It is an AP because difference between consecutive terms is equal.
=$$\sqrt{2},2\sqrt{2},3\sqrt{2},4\sqrt{2}$$
Common difference (d) = $$\sqrt{2}$$
Fifth term = $$4\sqrt{2} + \sqrt{2} = 5\sqrt{2}$$
Sixth term = $$5\sqrt{2} + \sqrt{2} = 6\sqrt{2}$$
Seventh term = $$6\sqrt{2} + \sqrt{2} = 7\sqrt{2}$$
Therefore, next three terms are $$5\sqrt{2}, 6\sqrt{2}, 7\sqrt{2}$$

(xiii) $$\sqrt{3}, \sqrt{6}, \sqrt{9}, \sqrt{12}$$…
It is not an AP because difference between consecutive terms is not equal.
= $$\sqrt{6} - \sqrt{3} \ne \sqrt{9} - \sqrt{6}$$

(xiv) $$1^2, 3^2, 5^2, 7^2$$…
It is not an AP because difference between consecutive terms is not equal.
= $$3^2 - 1^2 \ne 5^2 - 3^2$$

(xv) $$1^2, 5^2, 7^2, 73$$…
1, 25, 49, 73…
It is an AP because difference between consecutive terms is equal.
= $$5^2 - 1^2 = 7^2 - 5^2 = 73 - 7^2 = 24$$
Common difference (d) = 24
Fifth term = 73 + 24 = 97
Sixth term = 97 + 24 = 121
Seventh term = 121 + 24 = 145
Therefore, next three terms are 97, 121 and 145

#### Solution for Exercise 5.2

1. Find the missing variable from $$a, d, n$$ and $$a_n$$, where a is the first term, d is the common difference and $$a_n$$ is the nth term of AP.
(i) $$a = 7, d = 3, n = 8$$
(ii) $$a = –18, n = 10, a_n = 0$$
(iii) $$d = –3, n = 18, a_n = -5$$
(iv) $$a = –18.9, d = 2.5, a_n = 3.6$$
(v) $$a = 3.5, d = 0, n = 105$$

(i) a = 7, d = 3, n = 8
We need to find $$a_n$$ here.
Using formula $$a_n = a + (n - 1)d$$
Putting values of a, d and n,
$$a_n$$ = 7 + (8 - 1) 3
= 7 + (7) 3 = 7 + 21 = 28

(ii) a = –18, n= 10, $$a_n = 0$$
We need to find d here.
Using formula $$a_n = a + (n - 1)d$$
Putting values of a,$$a_n , n$$
0 = –18 + (10 – 1) d
0 = -18 + 9d
18 = 9d
=> d = 2

(iii) d = –3, n = 18, $$a_n = -5$$
We need to find a here.
Using formula $$a_n = a + (n - 1)d$$
Putting values of d, $$a_n , n$$
–5 = a + (18 – 1) (–3)
- 5 = a + (17) (-3)
-5 = a – 51
=> a = 46

(iv) a = –18.9, d = 2.5, $$a_n = 3.6$$
We need to find n here.
Using formula $$a_n = a + (n - 1)d$$
Putting values of $$d, a_n , n$$
3.6 = –18.9 + (n – 1) (2.5)
3.6 + 18.9 = (n - 1)(2.5)
22.5 = (n -1)(2.5)
(n - 1) = $$\frac{22.5}{2.5}$$
n - 1 = 9
=> n = 10

(v) a = 3.5, d = 0, n = 105
We need to find $$a_n$$ here.
Using formula $$a_n = a + (n - 1)d$$
Putting values of d, n and a,
$$a_n$$ = 3.5 + (105 - 1) (0)
=>$$a_n = 3.5 + 104 × 0$$
=> $$a_n = 3.5 + 0 => a_n = 3.5$$

2. Choose the correct choice in the following and justify:
(i) 30th term of the AP : $$10, 7, 4…....$$ is
(A) 97
(B) 77
(C) –77
(D) –87

(ii) 11th term of the AP : $$-3, \frac{1}{2}, 2…...$$ is
(A) 28
(B) 22
(C) –38
(D) $${-48}{\dfrac{1}{2}}$$

(i) 10, 7, 4…
First term = a = 10, Common difference = d = 7 – 10 = 4 – 7 = –3
And n = 30{Because, we need to find 30th term}
$$a_n = a + (n - 1)d$$
- $$a_{30}$$ = 10 + (30 - 1) (-3) = 10 – 87 = -77

(ii) -3, $${{-1} \over {2}}$$ , 2…
First term = a = –3, Common difference = d = $${{-1} \over {2}} - (-3) = 2 - ({{-1} \over {2}}) = {{5} \over {2}}$$
And n = 11 (Because, we need to find 11th term)
= -3 + (11 – 1) $${{5} \over {2}}$$= -3 + 25 = 22
Therefore 11th term is 22 which means answer is (B).

3. In the following AP’s find the missing terms:
(i) 2, __ , 26
(ii) __, 13, __, 3
(iii) 5, __, __, $${9}{\dfrac{1}{2}}$$
(iv) –4. __, __, __, __, 6
(v) __, 38, __, __, __, –22

(i) 2, __ , 26
We know that difference between consecutive terms is equal in any A.P.
Let the missing term be x.
x – 2 = 26 – x
2x = 28
=> x = 14
Therefore, missing term is 14.

(ii)__, 13, __, 3
Let missing terms be x and y.
The sequence becomes x, 13, y, 3
We know that difference between consecutive terms is constant in any A.P.
y – 13 = 3 – y
2y = 16
=> y = 8
And 13 – x = y – 13
x + y = 26
But, we have y = 8,
x + 8 = 26
=> x = 18
Therefore, missing terms are 18 and 8.

(iii) 5, __, __, $${9}{\dfrac{1}{2}}$$
Here, first term = a = 5 And, 4th term = $$a_4 = {9}{\dfrac{1}{2}}$$
Using formula $$a_n = a + (n - 1)d$$ , to find nth term of arithmetic progression,
$$a_4$$ = 5 + (4 - 1) d
$${{19} \over {2}}$$ = 5 + 3d
19 = 2 (5 + 3d)
19 = 10 + 6d
6d = 19 – 10
6d = 9
=> d = $${{3} \over {2}}$$
Therefore, we get common difference = d = $${{3} \over {2}}$$
Second term = a + d = $$5 + {{3} \over {2}} = {{13} \over {2}}$$
Third term = second term + d = $${{13} \over {2}} + {{3} \over {2}} = {{16} \over {2}} = 8$$
Therefore, missing terms are $${{13} \over {2}}$$ and 8

(iv)–4. __, __, __, __, 6
Here, First term = a = –4 and 6th term = $$a_6 = 6$$
Using formula $$a_n = a + (n - 1)d$$ , to find nth term of arithmetic progression,
$$a_6$$ = -4 + (6 - 1) d
6 = -4 + 5d
5d = 10
d = 2
Therefore, common difference = d = 2
Second term = first term + d = a + d = –4 + 2 = –2
Third term = second term + d = –2 + 2 = 0
Fourth term = third term + d = 0 + 2 = 2
Fifth term = fourth term + d = 2 + 2 = 4
Therefore, missing terms are –2, 0, 2 and 4.

(v) __, 38, __, __, __, –22
We are given 2nd and 6th term.
Using formula $$a_n = a + (n - 1)d$$ , to find nth term of arithmetic progression,
$$a_2$$ = a + (2 - 1) d And a6 = a + (6 - 1) d
38 = a + d And -22 = a + 5d
These are equations in two variables, we can solve them using any method.
Using equation (38 = a + d), we can say that a = 38 - d.
Putting value of a in equation (-22 = a + 5d),
-22 = 38 – d + 5d
4d = -60
d = -15
Using this value of d and putting this in equation 38 = a + d,
38 = a – 15- a = 53
Therefore, we get a = 53 and d = -15
First term = a = 53
Third term = second term + d = 38 – 15 = 23
Fourth term = third term + d = 23 – 15 = 8
Fifth term = fourth term + d = 8 – 15 = –7
Therefore, missing terms are 53, 23, 8 and –7.

4. Which term of the AP: 3, 8, 13, 18 … is 78?

First term = a = 3,
Common difference = d = 8 – 3 = 13 – 8 = 5
$$a_n$$ = 78
Using formula $$a_n = a + (n - 1)d$$, to find $$n^{th}$$ term of arithmetic progression,
78 = 3 + (n - 1) 5
75 = 5n - 5
80 = 5n
=> n = 16
It means $$16^{th}$$ term of the given AP is equal to 78.

5. Find the number of terms in each of the following APs:
(i) 7, 13, 19…., 205
(ii) 18,$${15}{\dfrac{1}{2}}$$ , 13…, -47

(i) 7, 13, 19 …, 205
First term = a = 7, Common difference = d = 13 – 7 = 19 – 13 = 6
And $$a_n = 205$$
Using formula $$a_n = a + (n - 1)d$$ , to find nth term of arithmetic progression,
205 = 7 + (n - 1) 6 = 7 + 6n – 6
205 = 6n + 1
204 = 6n
=> n = 34
Therefore, there are 34 terms in the given arithmetic progression.

(ii) 18, $${15}{\dfrac{1}{2}}$$ , 13 …, -47
First term = a =18, Common difference = d = $${15}{\dfrac{1}{2}} - 18 = {{31} \over {2}} - 18 = {{31 - 36} \over {2}} = {{-5} \over {2}}$$
And $$a_n = -47$$
Using formula $$a_n = a + (n - 1)d$$, to find nth term of arithmetic progression,
47 = 18 + (n - 1) $${{-5} \over {2}}$$
= 36 - $${{5} \over {2}}$$ n + $${{5} \over {2}}$$
-94 = 36 - 5n + 5
5n = 135
=> n = 27
Therefore, there are 27 terms in the given arithmetic progression.

6. Check whether –150 is a term of the AP: 11, 8, 5, 2…

Let -150 is the $$n^{th}$$ of AP 11, 8, 5, 2… which means that , $$a_n = -150$$
Here, First term = a = 11, Common difference = d = 8 – 11 = –3
Using formula $$a_n = a + (n - 1)d$$, to find $$n^{th}$$ term of arithmetic progression,
-150 = 11 + (n - 1) (-3)
-150 = 11 - 3n + 3
3n = 164
=> n = $${{164} \over {3}}$$
But, n cannot be in fraction.
So, -150 is not a term of this A.P.

7. Find the 31st term of an AP whose $$11^{th}$$ term is 38 and $$16^{th}$$ term is 73.

Given $$a_{11} = 38$$ and $$a_{16} = 73$$
Using formula $$a_n = a + ( n - 1)d$$, to find nth term of arithmetic progression,
38 = a + (11 - 1)(d), and
73 = a + (16 - 1)(d)
38 = a + 10d, and
73 = a + 15d
These are equations consisting of two variables.
We have, 38 = a + 10d
a = 38 - 10d
Let us put value of a in equation (73 = a + 15d),
73 = 38 - 10d + 15d
35 = 5d
Therefore, Common difference = d = 7
Putting value of d in equation 38 = a + 10d,
38 = a + 70
a = -32
Therefore, common difference = d = 7 and First term = a = –32
Using formula $$a_n = a + ( n - 1)d$$, to find nth term of arithmetic progression,
$$a_{31}$$= -32 + (31 - 1) (7)
= -32 + 210 = 178
Therefore, 31st term of AP is 178.

8. An AP consists of 50 terms of which $$3^{rd}$$ term is 12 and the last term is 106. Find the $$29^{th}$$ term.

An AP consists of 50 terms and the $$50^{th}$$ term is equal to 106 and $$a_3 = 12$$
Using formula $$a_n = a + (n - 1)d$$ , to find nth term of arithmetic progression,
$$a_{50}$$= a + (50 - 1)d, and
$$a_3$$= a + (3 - 1)d
106 = a + 49d and,
12 = a + 2d
These are equations consisting of two variables.
Using equation 106 = a + 49d, we get a = 106 - 49d
Putting value of a in the equation 12 = a + 2d,
12 = 106 - 49d + 2d
47d = 94
=> d = 2
Putting value of d in the equation, a = 106 - 49d,
a = 106 – 49 (2) = 106 – 98 = 8
Therefore, First term = a = 8 and Common difference = d = 2
To find $$29^{th}$$ term, we use formula $$a_n = a + (n - 1)d$$ which is used to find nth term of arithmetic progression,
$$a_{29}$$ = 8 + (29 - 1) 2 = 8 + 56 = 64
Therefore, 29th term of AP is equal to 64.

9. If the $$3^{rd}$$ and the $$9^{th}$$ terms of an AP are 4 and –8 respectively, which term of this AP is zero?

It is given that $$3^{rd}$$ and $$9^{th}$$ term of AP are 4 and –8 respectively.
It means $$a_3 = 4$$ and $$a_9 = -8$$
Using formula $$a_n = a + (n - 1)d$$, to find $$n^{th}$$ term of arithmetic progression,
4 = a + (3 – 1)d and,
–8 = a + (9 – 1)d
- 4 = a + 2d and,
-8 = a + 8d
These are equations in two variables.
Using equation 4 = a + 2d, we can say that a = 4 - 2d
Putting value of a in other equation -8 = a + 8d,
-8 = 4 - 2d + 8d
-12 = 6d
=> d = -2
Putting value of d in equation -8 = a + 8d,
-8 = a + 8 (-2)
-8 = a – 16
=> a = 8
Therefore, first term = a = 8 and Common Difference = d = -2
We want to know which term is equal to zero.
Using formula $$a_n = a + (n - 1)d$$ , to find $$n^{th}$$ term of arithmetic progression,
0 = 8 + (n - 1) (-2)
0 = 8 - 2n + 2
0 = 10 - 2n
2n = 10
=> n = 5
Therefore, $$5^{th}$$ term is equal to 0.

10. The $$17^{th}$$ term of an AP exceeds its $$10^{th}$$ term by 7. Find the common difference.

$$a_{17} = a_{10} + 7$$
Using formula $$a_n = a + (n - 1)d$$, to find nth term of arithmetic progression,
$$a_{17}$$ = a + 16d… (2)
$$a_{10}$$ = a + 9d… (3)
Putting (2) and (3) in equation (1),
a + 16d = a + 9d + 7
7d = 7
=> d = 1

11. Which term of the AP: 3, 15, 27, 39… will be 132 more than its $$54^{th}$$ term?

Lets first calculate $$54^{th}$$ of the given AP.
First term = a = 3, Common difference = d = 15 – 3 = 12
Using formula $$a_n = a + (n - 1)d$$, to find nth term of arithmetic progression,
$$a_{54}$$= a + (54 - 1) d = 3 + 53 (12) = 3 + 636 = 639
We want to find which term is 132 more than its $$54^{th}$$ term.
Let us suppose it is nth term which is 132 more than $$54^{th}$$ term.
$$a_n = a_{54} + 132$$
3 + (n - 1) 12 = 639 + 132
3 + 12n – 12 = 771
12n – 9 = 771
12n = 780
=> n = 65
Therefore, 65th term is 132 more than its $$54^{th}$$ term.

12. Two AP’s have the same common difference. The difference between their $$100^{th}$$ terms is 100, what is the difference between their $$1000^{th}$$ terms.

Let first term of $$1^{st}$$ AP = a
Let first term of $$2^{nd}$$ AP = a'
It is given that their common difference is same.
Let their common difference be d.
It is given that difference between their $$100^{th}$$ terms is 100.
Using formula $$a_n = a + (n - 1)d$$, to find nth term of arithmetic progression,
a + (100 - 1) d – [a' + (100 - 1) d]
= a + 99d - a' - 99d = 100
a - a' = 100… (1)
We want to find difference between their $$1000^{th}$$ terms which means we want to calculate:
a + (1000 - 1) d – [a' + (1000 - 1) d]
= a + 999d - a' - 999d = a – a'
Putting equation (1) in the above equation,
a + (1000 - 1) d – [a' + (1000 - 1) d]
= a + 999d - a' + 999d = a - a' = 100
Therefore, difference between their $$1000^{th}$$ terms would be equal to 100.

13. How many three digit numbers are divisible by 7?

We have AP starting from 105 because it is the first three digit number divisible by 7.
AP will end at 994 because it is the last three digit number divisible by 7.
Therefore, we have AP of the form 105, 112, 119…, 994
Let 994 is the $$n^{th}$$ term of AP.
We need to find n here.
First term = a = 105, Common difference = d = 112 – 105 = 7
Using formula $$a_n = a + (n - 1)d$$, to find $$n^{th}$$ term of arithmetic progression,
994 = 105 + (n - 1) (7)
994 = 105 + 7n - 7
896 = 7n
=> n = 128
It means 994 is the $$128^{th}$$ term of AP.
Therefore, there are 128 terms in AP.

14. How many multiples of 4 lie between 10 and 250?

First multiple of 4 which lie between 10 and 250 is 12.
The last multiple of 4 which lie between 10 and 250 is 248.
Therefore, AP is of the form 12, 16, 20… ,248
First term = a = 12, Common difference = d = 4
Using formula $$a_n$$ = a + (n – 1) d, to find $$n^{th}$$ term of arithmetic progression,
248 = 12 + (n - 1) (4)
248 = 12 + 4n - 4
240 = 4n
=> n = 60
It means that 248 is the $$60^{th}$$ term of AP.
So, we can say that there are 60 multiples of 4 which lie between 10 and 250.

15. For what value of n, are the nth terms of two AP’s: 63, 65, 67… and 3, 10, 17… equal?

Lets first consider AP 63, 65, 67…
First term = a = 63, Common difference = d = 65 – 63 = 2
Using formula , to find $$n^{th}$$ term of arithmetic progression,
$$a_n$$ = 63 + (n - 1) (2)… (1)
Now, consider second AP 3, 10, 17…
First term = a = 3, Common difference = d = 10 – 3 = 7
Using formula , to find $$n^{th}$$ term of arithmetic progression,
$$a_n$$ = 3 + (n - 1) (7)… (2)
According to the given condition:
(1) = (2)
63 + (n - 1) (2) = 3 + (n - 1) (7)
63 + 2n – 2 = 3 + 7n - 7
65 = 5n
=> n = 13
Therefore, $$13^{th}$$ terms of both the AP’s are equal.

16. Determine the AP whose third term is 16 and the $$7^{th}$$ term exceeds the $$5^{th}$$ term by 12.

Let first term of AP = a
Let common difference of AP = d
It is given that its 3rd term is equal to 16.
Using formula $$a_n = a + (n - 1)d$$, to find $$n^{th}$$ term of arithmetic progression,
16 = a + (3 - 1) (d)
16 = a + 2d… (1)
It is also given that $$7^{th}$$ term exceeds $$5^{th}$$ term by12.
According to the given condition:
$$a_7 = a_5 + 12$$
a + (7 – 1) d = a + (5 – 1) d + 12
2d = 12
=> d = 6
Putting value of d in equation 16 = a + 2d,
16 = a + 2(6)
=> a = 4
Therefore, first term = a = 4
And, common difference = d = 6
Therefore, AP is 4, 10, 16, 22…

17. Find the $$20^{th}$$ term from the last term of the AP: 3, 8, 13… , 253.

We want to find $$20^{th}$$ term from the last term of given AP.
So, let us write given AP in this way:253 … 13, 8, 3
Here First term = a = 253, Common Difference = d = 8 – 13 = –5
Using formula $$a_n = a + (n - 1)d$$, to find $$n^{th}$$ term of arithmetic progression,
= 253 + (20 - 1) (-5)
= 253 + 19 (-5) = 253 – 95 = 158
Therefore, the $$20^{th}$$ term from the last term of given AP is 158.

18. The sum of the $$4^{th}$$ and $$8^{th}$$ terms of an AP is 24 and the sum of $$6^{th}$$ and $$10^{th}$$ terms is 44. Find the three terms of the AP.

The sum of$$4^{th}$$ and $$8^{th}$$ terms of an AP is 24 and sum of $$6^{th}$$ and $$10^{th}$$ terms is 44.
$$a_4 + a_8 = 24$$ ,and $$a_6 + a_{10} = 44$$
Using formula $$a_n = a + (n - 1)d$$, to find nth term of arithmetic progression,
a + (4 - 1) d + [a + (8 - 1) d] = 24 and, a + (6 - 1) d + [a + (10 - 1) d] = 44
a + 3d + a + 7d = 24, and a + 5d + a + 9d = 44
2a + 10d = 24, and 2a + 14d = 44
a + 5d = 12, and a + 7d =22
These are equations in two variables.
Using equation, a + 5d = 12, we can say that a = 12 - 5d…...(1)
Putting (1) in equation a + 7d = 22,
12 - 5d + 7d = 22
12 + 2d = 22
2d = 10
=> d = 5
Putting value of d in equation a = 12 - 5d,
a = 12 – 5 (5) = 12 – 25 = -13
Therefore, first term = a = -13 and, Common difference = d = 5
Therefore, AP is –13, –8, –3, 2…
Its first three terms are –13, –8 and –3.

19. Subba Rao started work in 1995 at an annual salary of Rs 5000 and received an increment of Rs 200 each year. In which year did his income reach Rs 7000?

Subba Rao’s starting salary = Rs 5000
It means, first term = a = 5000
He gets an increment of Rs 200 after every year.
Therefore, common difference = d = 200
His salary after 1 year = 5000 + 200 = Rs 5200
His salary after two years = 5200 + 200 = Rs 5400
Therefore, it is an AP of the form 5000, 5200, 5400, 5600… , 7000
We want to know in which year his income reaches Rs 7000.
Using formula $$a_n = a + (n - 1)d$$ , to find $$n^{th}$$ term of arithmetic progression,
7000 = 5000 + (n - 1) (200)
7000 = 5000 + 200n - 200
7000 – 5000 + 200 = 200n
2200 = 200n
n = 11
It means after 11 years, Subba Rao’s income would be Rs 7000.

20. Ramkali saved Rs. 5 in the first week of a year and then increased her weekly savings by Rs. 1.75. If in the nth week, her weekly savings become Rs 20.75, find n.

Ramkali saved Rs. 5 in the first week of year. It means first term = a = 5
Ramkali increased her weekly savings by Rs 1.75.
Therefore, common difference = d = Rs 1.75
Money saved by Ramkali in the second week = a + d = 5 + 1.75 = Rs 6.75
Money saved by Ramkali in the third week = 6.75 + 1.75 = Rs 8.5
Therefore, it is an AP of the form:5, 6.75, 8.5 … , 20.75
We want to know in which year her weekly savings become 20.75.
Using formula $$a_n = a + (n - 1)d$$, to find $$n^{th}$$ term of arithmetic progression,
20.75 = 5 + (n - 1) (1.75)
20.75 = 5 + 1.75n - 1.75
17.5 = 1.75n
=> n = 10
It means in the $$10^{th}$$ week her savings become Rs 20.75.

#### Solution for Exercise 5.3

1. Find the sum of the following AP’s.
(i) 2, 7, 12… to 10 terms
(ii) –37, –33, –29… to 12 terms
(iii) 0.6, 1.7, 2.8… to 100 terms
(iv) $${{1} \over {15}},{{1} \over {12}},{{1} \over {10}}…$$to 11 terms

The formula for finding the sum of A.P. is $$S_n = {{n} \over {2}}[2a + (n - 1)d]$$
Where a= first term, d = common difference and n = number of terms
(i)2, 7, 12… to 10 terms
Here a = 2, d = $$7 - 2 = 5$$ and n = 10
$$S_n = {{10} \over {2}}(4 + (10 - 1)5) = 5(4 + 45) = 5(49) = 245$$

(ii) –37, –33, –29… to 12 terms
Here a = -37, d = $$-33 - (-37) = 4$$ and n = 12
$$S_n = {{12} \over {2}}(-74 + (12 - 1)4) = 6(-74 + 44) = 6(-30) = -180$$

(iii) 0.6, 1.7, 2.8… to 100 terms
Here a = 0.6, d = $$1.7 - 0.6 = 1.1$$ and n = 100
$$S_n = {{100} \over {2}}(1.2 + (100 - 1)1.1) = 50(1.2 + 108.9) = 50(110.1) = 5505$$

(iv) $${{1} \over {15}},{{1} \over {12}},{{1} \over {10}}…$$to 11 terms
Here a = $${{1} \over {15}}$$, d = $${{1} \over {12}} - {{1} \over {15}} = {{1} \over {60}}$$ and n = 100
$$S_n = {{11} \over {2}}({{2} \over {15}} + (11 - 1){{1} \over {60}}) = {{11} \over {2}}({{2} \over {15}} + {{1} \over {6}}) = {{11} \over {2}}({{9} \over {30}}) = {{33} \over {20}}$$

2. Find the sums given below:
(i) $$7 + {10}{\dfrac{1}{2}} + 14 + .......+ 84$$
(ii) 34 + 32 + 30 + .....… + 10
(iii) –5 + (–8) + (–11) + …..... + (–230)

(i) $$7 + {10}{\dfrac{1}{2}} + 14 +..........+ 84$$
Here First term = a = 7, Common difference = d = $${{21} \over {2}} - 7 = {{21 - 15} \over {2}} = {{7} \over {2}} = 3.5$$
And Last term = l = 84
We do not know how many terms are there in the given AP.
So, we need to find n first.
Using formula $$a_n = a + (n - 1)d$$ , to find nth term of arithmetic progression,
[7 + (n - 1) (3.5)] = 84
7 + (3.5) n - 3.5 = 84
3.5n = 84 + 3.5 – 7
3.5n = 80.5
n = 23
Therefore, there are 23 terms in the given AP.
It means n = 23.
Applying formula $$S_n = {{n} \over {2}} (a + l)$$, to find sum of n terms of AP,
$$S_n = {{23} \over {2}} (7 + 84)$$
$$S_{23} = {{23} \over {2}} (91) = 1046.5$$

(ii) 34 + 32 + 30 + … + 10
Here First term = a = 34, Common difference = d = 32 – 34 = –2
And Last term = l = 10
We do not know how many terms are there in the given AP.
So, we need to find n first.
Using formula $$a_n = a + (n - 1)d$$, to find nth term of arithmetic progression,
[34 + (n - 1) (-2)] = 10
34 – 2n + 2 = 10
2n = 26
=> n = 13
Therefore, there are 13 terms in the given AP.
It means n = 13.
Applying formula $$S_n = {{n} \over {2}} (a + l)$$, to find sum of n terms of AP,
$$S_{13} = {{13} \over {2}} (34 + 10) = {{13} \over {2}} (44) = 286$$

(iii) -5 + (-8) + (-11) + … + (-230)
Here First term = a = –5, Common difference = d = –8 – (–5) = –8 + 5 = –3
And Last term = l = -230
We do not know how many terms are there in the given AP.
So, we need to find n first.
Using formula $$a_n = a + (n - 1)d$$, to find nth term of arithmetic progression,
[-5 + (n - 1) (-3)] = -230
-5 - 3n + 3 = -230
-3n = -228
=> n = 76
Therefore, there are 76 terms in the given AP.
It means n = 76.
Applying formula $$S_n = {{n} \over {2}} (a + l)$$, to find sum of n terms of AP,
$$S_{76} = {{76} \over {2}} (-5 - 230) = 38 (-235) = -8930$$

3. In an AP
(i)Given a = 5, d = 3, an = 50, find n and $$S_n$$.
(ii) Given a = 7, $$a_{13}$$ = 35, find d and $$S_{13}$$.
(iii) Given $$a_{12}$$ = 37, d = 3, find a and $$S_{12}$$.
(iv) Given $$a_3$$ = 15, $$S_{10}$$ = 125, find d and $$a_{10}$$.
(v) Given d = 5, $$S_9$$= 75, find a and $$a_9$$.
(vi) Given a = 2, d = 8, $$S_n$$ = 90, find n and an.
(vii) Given a = 8, an = 62, $$S_n$$ = 210, find n and d.
(viii) Given an = 4, d = 2, $$S_n$$ = - 14, find n and a.
(ix) Given a = 3, n = 8, S = 192, find d.
(x)Given l = 28, S = 144 and there are total 9 terms. Find a.

(i) Given that, a = 5, d = 3, $$a_n$$ = 50
As we know, from the formula of the nth term in an AP,
$$a_n = a + (n – 1)d$$
Therefore, putting the given values, we get,
= 50 = 5+(n -1)×3
3(n -1) = 45
n -1 = 15
n = 16
Now, sum of n terms,
$$S_n = (a + a_n)$$
$$S_n = {{16} \over{2}} (5 + 50) = 440$$

(ii) Given that, a = 7, $$a_{13}$$ = 35
As we know, from the formula of the nth term in an AP
$$a_n = a + (n – 1)d$$
Therefore, putting the given values, we get,
= 35 = 7+(13-1)d
12d = 28
d = 28/12 = 2.33
Now, sum of n terms,
$$S_n = (a + a_n)$$
$$S_n = {{13} \over {2}} (7 + 35) = 273$$

(iii)Given that, $$a_{12}$$ = 37, d = 3
As we know, from the formula of the nth term in an AP,
$$a_n = a + (n – 1)d$$
Therefore, putting the given values, we get,
$$a_{12}$$ = a + (12 - 1)3
37 = a + 33
a = 4
Now, sum of nth term,
$$S_n = (a + a_n)$$
$$S_n = {{12} \over {2}} (4 + 37)$$
= 246

(iv) Given that, $$a_3$$ = 15, $$S_{10}$$ = 12
As we know, from the formula of the nth term in an AP,
$$a_n = a + (n – 1)d$$
Therefore, putting the given values, we get,
$$a_3$$ = a+(3 - 1)d
15 = a + 2d ………………………….. (i)
Sum of the nth term,
$$S_n = {{n} \over {2}} [2a+(n-1)d]$$
$$S_{10} = {{10} \over {2}} [2a+(10-1)d]$$
125 = 5(2a+9d)
25 = 2a+9d ……………………….. (ii)
On multiplying equation (i) by 2, we will get;
30 = 2a+4d ………………………………. (iii)
By subtracting equation (iii) from (ii), we get,
-5 = 5d
d = -1
From equation (i)
15 = a + 2(-1)
15 = a - 2
a = 17 = First term
$$a_{10}$$ = a + (10-1)d
$$a_{10}$$ = 17 + (9)(-1)
$$a_{10}$$ = 17 -9 = 8

(v) Given that, d = 5, $$S_9$$ = 75
As, sum of n terms in AP is,
$$S_n = {{n} \over {2}} [2a+(n-1)d]$$
Therefore, the sum of first nine terms are;
$$S_9 = {{9} \over {2}} [2a+(9 -1)5]$$=25 = 3(a+20)
25 = 3a+60
3a = 25-60
a = $${{-35} \over {3}}$$
As we know, the nth term can be written as;
$$a_n = a + (n – 1)d$$
$$a_9 = a + (9 - 1)(5)$$
= $${{-35} \over {3}}$$ +8(5)
= $${{-35} \over {3}}$$+40
= ($${{-35 + 120} \over {3}}$$) = $${{85} \over {3}}$$

(vi) Given that, a = 2, d = 8, $$S_n$$ = 90
As, sum of n terms in an AP is,
$$S_n = {{n} \over {2}} [2a+(n-1)d]$$
$$90 = {{n} \over {2}} [2a +(n -1)d]$$
=> $$180 = n(4 +8n - 8)$$
$$180 = n(8n-4) = 8n^2-4n$$
$$8n^2 - 4n –180 = 0$$
$$2n^2 - n - 45 = 0$$
$$2n^2 -10n + 9n - 45 = 0$$
$$2n(n - 5) + 9(n - 5) = 0$$
$$(2n - 9)(2n + 9) = 0$$
So, n = 5 (as it is positive integer)
$$a_5$$ = 8 + 5×4 = 34

(vii) Given that, a = 8, $$a_n$$ = 62, $$S_n$$ = 210
As, sum of n terms in an AP is,
$$S_n = {{n} \over {2}} [2a+(n-1)d]$$
$$210 = {{n} \over {2}} (8 +62)$$
35n = 210
n = $$\frac{210}{35}$$ = 6
Now, 62 = 8 + 5d
5d = 62 - 8 = 54
d = $${{54} \over {5}}$$ = 10.8

(viii) Given that, nth term, an = 4, common difference, d = 2, sum of n terms, $$S_n$$ = -14.
As we know, from the formula of the nth term in an AP,
$$a_n = a + (n – 1)d$$
Therefore, putting the given values, we get,
4 = a+(n -1)2
4 = a+2n-2
a+2n = 6
a = 6 - 2n …………………………………………. (i)
As we know, the sum of n terms is;
$$S_n = \frac{n}{2} (a + a_n)$$
$$-14 = \frac{n}{2}(a+4)$$
$$-28 = n(a + 4)$$
$$-28 = n(6 -2n + 4)$${From equation (i)}
$$-28 = n(-2n +10)$$
$$-28 = - 2n^2 + 10n$$
$$2n^2 -10n - 28 = 0$$
$$n^2 -5n -14 = 0$$
$$n^2 -7n + 2n -14 = 0$$
$$n(n - 7) + 2(n - 7) = 0$$
$$(n -7)(n +2) = 0$$
Either n - 7 = 0 or n + 2 = 0
n = 7 or n = -2
However, n can neither be negative nor fractional.
Therefore, n = 7
From equation (i), we get
a = 6 - 2n
a = 6 - 2(7)
= 6 - 14
Thus, a = -8

(ix) Given that, first term, a = 3,
Number of terms, n = 8
And sum of n terms, S = 192
As we know,
$$S_n = {{n} \over {2}} [2a+(n-1)d]$$
$$192 = {{8} \over {2}} [2×3+(8 -1)d]$$
192 = 4[6 + 7d]
48 = 6 + 7d
42 = 7d
d = 6

(x) Given that, l = 28,S = 144 and there are total of 9 terms.
Sum of n terms formula,
$$S_n = {{n} \over {2}} (a + l)$$
$$144 = {{9} \over {2}}(a+28)$$
(16)×(2) = a + 28
32 = a + 28
a = 4

4. How many terms of the AP. 9, 17, 25 … must be taken to give a sum of 636?

Let there be n terms of the AP. 9, 17, 25 …
For this A.P.,
First term, a = 9
Common difference, d = $$a_2 – a_1$$ = 17-9 = 8
As, the sum of n terms, is;
$$S_n = {{n} \over {2}} (2a + (n – 1)d$$
636 = $${{n} \over {2}}$$ [2×a+(8-1)×8]
636 = $${{n} \over {2}}$$ [18+(n-1)×8]
636 = n [9 +4n -4]
636 = n (4n +5)
4$$n^2$$ +5n -636 = 0
4$$n^2$$ +53n -48n -636 = 0
n (4n + 53)-12 (4n + 53) = 0
(4n +53)(n -12) = 0
Either 4n+53 = 0 or n-12 = 0
n = $${{-53} \over {4}}$$ or n = 12
n cannot be negative or fraction, therefore, n = 12 only.

5. The first term of an AP is 5, the last term is 45 and the sum is 400. Find the number of terms and the common difference.

Given that,
first term, a = 5
last term, l = 45
Sum of the AP, $$S_n$$ = 400
As we know, the sum of AP formula is;
$$S_n = {{n} \over {2}} (a+l)$$
400 = $${{n} \over {2}}$$(5+45)
400 = $${{n} \over {2}}$$(50)
Number of terms, n =16
As we know, the last term of AP series can be written as;
l = a+(n -1)d
45 = 5 +(16 -1)d
40 = 15d
Common difference, d = $${{40} \over {15}} = {{8} \over {3}}$$

6. The first and the last term of an AP are 17 and 350 respectively. If the common difference is 9, how many terms are there and what is their sum?

Given that,
First term, a = 17
Last term, l = 350
Common difference, d = 9
Let there be n terms in the A.P., thus the formula for last term can be written as;
l = a+(n - 1)d
350 = 17+(n - 1)9
333 = (n -1)9
(n- 1) = 37
n = 38
$$S_n = {{n} \over {2}} (a+l)$$
$$S_{38} = {{13} \over {2}} (17+350)$$
= 19×367
= 6973
Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973

7. Find the sum of first 22 terms of an AP in which d = 7 and $$22^{nd}$$ term is 149.

Given,
Common difference, d = 7
$$22^{nd}$$ term, $$a_{22}$$ = 149
We have to find the sum of first 22 term, $$S_{22}$$
By the formula of nth term,
$$a_n = a + (n - 1)d$$
$$a_{22} = a + (22 - 1)d$$
149 = a + (21)(7)
149 = a + 147
a = 2 = First term
Sum of n terms,
$$S_n = {{n} \over {2}}(a+a_n)$$ = $${{22} \over {2}}$$ (2+149) = 11×151 = 1661

8. Find the sum of first 51 terms of an AP whose second and third terms are 14 and 18 respectively.

Given that,
Second term, $$a_2$$ = 14
Third term, $$a_3$$ = 18
Common difference, d = $$a_3 - a_2$$ = 18-14 = 4
$$a_2$$ = a + d
14 = a + 4
a = 10 = First term
Sum of n terms;
$$S_n = {{n} \over {2}} [2a + (n – 1)d]$$
$$S_{51} = {{51} \over {2}} [2×10 (51-1) 4]$$
= $${{51} \over {2}}[2 + (20)4$$]
= 51×$${{220} \over {2}}$$
= 51×110
= 5610

9. If the sum of first 7 terms of an AP is 49 and that of 17 terms is 289, find the sum of first n terms.

Given that,
$$S_7$$ = 49
$$S_{17}$$ = 289
We know, Sum of nth term;
$$S_n = {{n} \over {2}} [2a + (n – 1)d]$$
Therefore,
$$S_7= {{7} \over {2}} [2a + (n -1)d]$$
$$S_7 = {{7} \over {2}} [2a + (7 -1)d]$$
$$49 ={{7} \over {2}} [2a + 6d]$$
7 = (a+3d)
a + 3d = 7 …………………………………. (i)
In the same way,
$$S_{17} = {{17} \over {2}} [2a + (17-1)d]$$
$$289 = {{17} \over {2}} (2a + 16d)$$
17 = (a + 8d)
a + 8d = 17 ………………………………. (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i), we can write it as;
a + 3(2) = 7
a + 6 = 7
a = 1
Hence, $$S_n = {{n} \over {2}} [2a + (n – 1)d]$$
= $${{n} \over {2}}[2(1)+(n – 1)×2]$$ = $${{n} \over {2}}(2+2n-2)$$ = $${{n} \over {2}}(2n)$$ = $$n^2$$

10. Show that $$a_1, a_2 … , a_n , …$$ form an AP where an is defined as below
(i) $$a_n = 3+4_n$$
(ii) $$a_n = 9 - 5_n$$
Also find the sum of the first 15 terms in each case.

(i) $$a_n = 3+4_n$$
$$a_1 = 3+4(1) = 7$$
$$a_2 = 3+4(2) = 3+8 = 11$$
$$a_3 = 3+4(3) = 3+12 = 15$$
$$a_4 = 3+4(4) = 3+16 = 19$$
We can see here, the common difference between the terms are;
$$a_2 - a_1 = 11-7 = 4$$
$$a_3 - a_2 = 15-11 = 4$$
$$a_4 - a_3 = 19-15 = 4$$
Hence, $$a_k + 1 - a_k$$ is the same value every time. Therefore, this is an AP with common difference as 4 and first term as 7.
Now, we know, the sum of $$n^{th}$$ term is;
$$S_n = {{n} \over {2}}[2a+(n -1)d]$$
$$S_{15} = {{15} \over {2}}[2(7)+(15-1)×4]$$
= $${{15} \over {2}}[(14)+56]$$
= $${{15} \over {2}}(70)$$
= 15×35
= 525

(ii) $$a_n = 9 - 5n$$
$$a_1 = 9-5×1 = 9-5 = 4$$
$$a_2 = 9-5×2 = 9-10 = -1$$
$$a_3 = 9-5×3 = 9-15 = -6$$
$$a_4 = 9-5×4 = 9-20 = -11$$
We can see here, the common difference between the terms are;
$$a_2 - a_1 = -1-4 = -5$$
$$a_3 - a_2 = -6-(-1) = -5$$
$$a_4 - a_3 = -11-(-6) = -5$$
Hence, $$a_k + 1 - a_k$$ is same every time. Therefore, this is an A.P. with common difference as -5 and first term as 4.
Now, we know, the sum of nth term is;
$$S_n = {{n} \over {2}} [2a +(n-1)d]$$
$$S_{15} = {{15} \over {2}}[2(4) +(15 -1)(-5)]$$
= $${{15} \over {2}}$$[8 + 14(-5)]
= $${{15} \over {2}}$$(8 - 70)
= $${{15} \over {2}}$$(-62)
= 15(-31)
= -465

11. If the sum of the first n terms of an AP is $$4n - n^2$$, what is the first term (that is $$S_1$$)? What is the sum of first two terms? What is the second term? Similarly find the $$3^{rd}$$, the $$10^{th}$$ and the nth terms.

Given that,
$$S_n = 4n-n^2$$
First term, a = $$S_1 = 4(1) - (1)2 = 4-1 = 3$$
Sum of first two terms = $$S_2= 4(2)-(2)2 = 8-4 = 4$$
Second term, $$a_2 = S_2 - S_1 = 4-3 = 1$$
Common difference, d = $$a_2-a = 1-3 = -2$$
Nth term, $$a_n = a+(n-1)d$$
= 3 + (n -1)(-2)
= 3 - 2n + 2
= 5 - 2n
Therefore,$$a_3 = 5 - 2(3) = 5 - 6 = -1$$
$$a_{10} = 5-2(10) = 5 - 20 = -15$$
Hence, the sum of first two terms is 4. The second term is 1.
The 3rd, the 10th, and the nth terms are -1, -15, and 5 - 2n respectively.

12. Find the sum of first 40 positive integers divisible by 6.

The positive integers that are divisible by 6 are 6, 12, 18, 24 ….
We can see here, that this series forms an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
We have to find $$S_{40}$$
By the formula of sum of n terms, we know,
$$S_n = {{n} \over {2}} [2a +(n – 1)d]$$
Therefore, putting n = 40, we get,
$$S_{40} = {{40} \over {2}} [2(6)+(40-1)6]$$
= $$20[12 + (39)(6)]$$
= $$20(12 + 234)$$
= $$20 × 246$$
= $$4920$$

13. Find the sum of first 15 multiples of 8.

The multiples of 8 are 8, 16, 24, 32…
The series is in the form of AP, having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
We have to find $$S_{15}$$
By the formula of sum of $$n^{th}$$ term, we know,
$$S_n = {{n} \over {2}} [2a+(n-1)d]$$
$$S_{15} = {{15} \over {2}} [2(8) + (15-1)8] = {{15} \over {2}}[6 +(14)(8)] = {{15} \over {2}}[16 +112] = {{15(128)} \over {2}} = 15 × 64 = 960$$

14. Find the sum of the odd numbers between 0 and 50.

The odd numbers between 0 and 50 are 1, 3, 5, 7, 9 … 49.
Therefore, we can see that these odd numbers are in the form of A.P.
Hence,
First term, a = 1
Common difference, d = 2
Last term, l = 49
By the formula of last term, we know,
$$l = a+(n-1) d =>49 = 1+(n-1)2 =>48 = 2(n -1)$$
n - 1 = 24
n = 25 = Number of terms
By the formula of sum of $$n^{th}$$ term, we know,
$$S_n = {{n} \over {2}}(a +l)$$
$$S_{25} = {{25} \over {2}} (1+49) = {{25(50)} \over {2}} =(25)(25) = 625$$

15. A contract on construction job specifies a penalty for delay of completion beyond a certain dates follows: Rs. 200 for the first day, Rs. 250 for the second day, Rs. 300 for the third day, etc., the penalty for each succeeding day being Rs. 50 more than for the preceding day. How much money the contractor has to pay as penalty, if he has delayed the work by 30 days.

We can see, that the given penalties are in the form of A.P. having first term as 200 and common difference as 50.
Therefore, a = 200 and d = 50
Penalty that has to be paid if contractor has delayed the work by 30 days = $$S_{30}$$
By the formula of sum of nth term, we know,
$$S_n = {{n} \over {2}}[2a+(n -1)d]$$
Therefore,
$$S_{30}= {{30} \over {2}}[2(200)+(30 – 1)50] = 15[400+1450] = 15(1850) = 27750$$
Therefore, the contractor has to pay Rs 27750 as penalty.

16. A sum of Rs 700 is to be used to give seven cash prizes to students of a school for their overall academic performance. If each prize is Rs 20 less than its preceding prize, find the value of each of the prizes.

Let the cost of 1st prize be P Rs.
Cost of 2nd prize = P - 20 Rs
And cost of 3rd prize = P - 40 Rs
We can see that the cost of these prizes are in the form of A.P., having common difference as -20 and first term as P.
Thus, a = P and d = -20
Given that, $$S_7 = 700$$
By the formula of sum of nth term, we know, $$S_n = {{n} \over {2}} [2a + (n – 1)d]$$
$${{7} \over {2}} [2a + (7 – 1)d] = 700$$
$${{2a + (6)(-20)} \over {2}} = 100$$
a + 3(-20) = 100
a - 60 = 100
a = 160
Therefore, the value of each of the prizes was Rs 160, Rs 140, Rs 120, Rs 100, Rs 80, Rs 60, and Rs 40.

17. In a school, students thought of planting trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be the same as the class, in which they are studying, e.g., a section of class I will plant 1 tree, a section of class II will plant 2 trees and so on till class XII. There are three sections of each class. How many trees will be planted by the students?

t can be observed that the number of trees planted by the students is in an AP.
1, 2, 3, 4, 5………………..12
First term, a = 1
Common difference, d = 2 - 1 = 1
$$S_n = {{n} \over {2}} [2a +(n-1)d]$$
$$S_{12} = {{12} \over {2}} [2(1)+(12-1)(1)] = 6(2+11) = 6(13) = 78$$
Therefore, number of trees planted by 1 section of the classes = 78
Number of trees planted by 3 sections of the classes = 3×78 = 234
Therefore, 234 trees will be planted by the students.

18. A spiral is made up of successive semicircles, with centres alternately at A and B, starting with centre at A of radii 0.5, 1.0 cm, 1.5 cm, 2.0 cm, ……… as shown in figure. What is the total length of such a spiral made up of thirteen consecutive semicircles? (Take $$\pi$$ = 22/7)

We know,
Perimeter of a semi-circle = $$\pi$$r
Therefore,
$$P_1 = \pi(0.5) = {{\pi} \over {2}}$$ cm
$$P_2 = \pi(1) = \pi$$ cm
$$P_3 = \pi(1.5) = {{3pi} \over { 2}}$$ cm
Where, $$P_1, P_2, P_3$$ are the lengths of the semi-circles.
Hence we got a series here, as,
$${{\pi} \over {2}} , \pi, {{3\pi} \over { 2}}, 2\pi,$$ ….
$$P_1 = {{\pi} \over {2}}$$ cm
$$P_2 = \pi$$ cm
Common difference, $$d = P_2- P_1 = \pi – {{\pi} \over {2}} = {{\pi} \over {2}}$$
First term = $$P_1= a = {{\pi} \over {2}}$$ cm
By the sum of n term formula, we know,
$$S_n = {{n} \over {2}} [2a + (n – 1)d]$$
Therefor, Sum of the length of 13 consecutive circles is;
$$S_{13} = {{13} \over {2}} [2({{\pi} \over {2}}) + (13 – 1){{\pi} \over {2}}]$$
= $$\frac{13}{2} [\pi + 6\pi]$$
=$$\frac{13}{2} (7\pi)$$
= $$\frac{13}{2} × 7 × \frac{22}{7}$$
= $$143$$ cm

19. 200 logs are stacked in the following manner: 20 logs in the bottom row, 19 in the next row, 18 in the row next to it and so on. In how many rows are the 200 logs placed and how many logs are in the top row?

We can see that the numbers of logs in rows are in the form of an A.P.20, 19, 18…
For the given A.P.,
First term, a = 20 and common difference,$$d = a_2 - a_1 = 19-20 = -1$$
Let a total of 200 logs be placed in n rows.
Thus, $$S_n = 200$$
By the sum of $$n^{th}$$ term formula,
$$S_n = {{n} \over {2}} [2a +(n -1)d]$$
$$S_{12} = {{12} \over {2}} [2(20)+(n -1)(-1)]$$
400 = n (40- n+1)
400 = n (41-n)
$$400 = 41n-n^2$$
$$n^2-41n + 400 = 0$$
$$n^2-16n-25n+400 = 0$$
n(n -16)-25(n -16) = 0
(n -16)(n -25) = 0
Either (n -16) = 0 or n-25 = 0
n = 16 or n = 25
By the nth term formula,
$$a_n = a+(n-1)d$$
$$a_{16} = 20+(16-1)(-1)$$
$$a_{16} = 20-15$$
$$a_{16} = 5$$
Similarly, the 25th term could be written as;
$$a_{25} = 20+(25-1)(-1)$$
$$a_{25} = 20-24$$
= -4
It can be seen, the number of logs in 16th row is 5 as the numbers cannot be negative.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16th row is 5.

20. In a potato race, a bucket is placed at the starting point, which is 5 m from the first potato and other potatoes are placed 3 m apart in a straight line. There are ten potatoes in the line.
A competitor starts from the bucket, picks up the nearest potato, runs back with it, drops it in the bucket, runs back to pick up the next potato, runs to the bucket to drop it in, and she continues in the same way until all the potatoes are in the bucket. What is the total distance the competitor has to run? [Hint: to pick up the first potato and the second potato, the total distance (in metres) run by a competitor is 2×5+2×(5+3)]

The distances of potatoes from the bucket are 5, 8, 11, 14…, which is in the form of AP.
Given, the distance run by the competitor for collecting these potatoes are two times of the distance at which the potatoes have been kept.
Therefore, distances to be run w.r.t distances of potatoes, could be written as;
10, 16, 22, 28, 34,……….
Hence, the first term, a = 10 and d = 16 - 10 = 6
S10 =?
By the formula of sum of n terms, we know,
$$S_{10} = {{12} \over {2}} [2(20)+(n -1)(-1)] = 5[20+54] = 5(74) = 370$$
Therefore, the competitor will run a total distance of 370 m.

#### Solution for Exercise 5.4

1. Which term of the AP : 121, 117, 113, . . ., is its first negative term? [Hint : Find n for an < 0]

Given the AP series is 121, 117, 113, . . .,
Thus, first term, a = 121
Common difference, d = 117-121= -4
By the $$n^{th}$$ term formula,
$$a_n = a+(n -1)d$$
Therefore,
$$a_n = 121+(n-1)(-4) = 121-4n+4 =125-4n$$
To find the first negative term of the series, $$a_n < 0$$
Therefore,
125-4n < 0
125 < 4n
n>125/4
n>31.25
Therefore, the first negative term of the series is $$32^{nd}$$ term.

2. The sum of the third and the seventh terms of an AP is 6 and their product is 8. Find the sum of first sixteen terms of the AP.

From the given statements, we can write,
$$a_3 + a_7 = 6$$ …………………………….(i)
And
$$a_3 ×a_7 = 8$$ ……………………………..(ii)
By the $$n^{th}$$ term formula,
$$a_n = a+(n - 1)d$$
Third term, $$a_3 = a+(3 -1)d$$
$$a_3 = a + 2d$$………………………………(iii)
And Seventh term, $$a_7= a+(7-1)d$$
$$a_7 = a + 6d$$………………………………..(iv)
From equation (iii) and (iv), putting in equation(i), we get,
a+2d +a+6d = 6
2a+8d = 6
a+4d=3
or
a = 3–4d …………………………………(v)
Again putting the eq.(iii) and (iv), in eq. (ii), we get,
(a+2d)×(a+6d) = 8
Putting the value of a from equation (v), we get,
$$(3–4d +2d)×(3–4d+6d) = 8$$
$$(3 –2d)×(3+2d) = 8$$
$$3^2 – 2d^2$$ = 8
$$9 – 4d^2$$ = 8
$$4d^2 = 1$$
$$d = {{1} \over {2}} or -{{1} \over {2}}$$
Now, by putting both the values of d, we get,
$$a = 3 – 4d = 3 – 4({{1} \over {2}}) = 3 – 2 = 1$$, when $$d = {{1} \over {2}}$$
$$a = 3 – 4d = 3 – 4(-{{1} \over {2}}) = 3+2 = 5$$, when $$d = -{{1} \over {2}}$$
We know, the sum of nth term of AP is;
$$S_n = {{n} \over {2}} [2a +(n – 1)d]$$
So, when a = 1 and d=$${{1} \over {2}}$$
Then, the sum of first 16 terms are;
$$S_{16} = {{16} \over {2}} [2 +(16-1){{1} \over {2}}] = 8(2+{{15} \over {2}}) = 76$$
And when a = 5 and d= -$${{1} \over {2}}$$
Then, the sum of first 16 terms are;
$$S_{16} = {{16} \over {2}} [2.5+(16-1)(-{{1} \over {2}})] = 8({{5} \over {2}})=20$$

3. A ladder has rungs 25 cm apart.. The rungs decrease uniformly in length from 45 cm at the bottom to 25 cm at the top. If the top and the bottom rungs are $${2} {\dfrac{1}{2}}$$ apart, what is the length of the wood required for the rungs? [Hint : Number of rungs = -250/25 ].

Given,
Distance between the rungs of the ladder is 25 cm.
Distance between the top rung and bottom rung of the ladder is =
$${2} {\dfrac{1}{2}}m = {2} {\dfrac{1}{2}} (100)cm = {{5} \over {2}} ×100$$ cm
= 250 cm
Therefore, total number of rungs = $${{250} \over {25}} + 1 = 11$$
As we can see from the figure, the ladder has rungs in decreasing order from top to bottom. Thus, we can conclude now, that the rungs are decreasing in an order of AP.
And the length of the wood required for the rungs will be equal to the sum of the terms of AP series formed.
So,
First term, a = 45
Last term, l = 25
Number of terms, n = 11
Now, as we know, sum of nth terms is equal to,
$$S_n= {{n} \over {2}}(a+ l)$$
$$S_n= {{11} \over {2}}(45+25) = {{11} \over {2}}(70) = 385$$ cm
Hence, the length of the wood required for the rungs is 385 cm.

4. The houses of a row are numbered consecutively from 1 to 49. Show that there is a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it. Find this value of x.
[Hint :Sx – 1 = S49 – Sx]

Given,
Row houses are numbers from 1,2,3,4,5…….49.
Thus we can see the houses numbered in a row are in the form of AP.
So,
First term, a = 1
Common difference, d=1
Let us say the number of $$x^{th}$$ houses can be represented as;
Sum of nth term of AP = $${{n} \over {2}}[2a + (n - 1)d]$$
Sum of number of houses beyond x house = Sx-1
= $${{x - 1} \over {2}}[2(1) + (x -1-1)1]$$
= $${{x - 1} \over {2}} [2+x-2]$$
= $${{x(x - 1)} \over {2}}$$………………………………………(i)
By the given condition, we can write,
$$S_{49} – Sx = ({{49} \over {2}}[2.1+(49-1)1])–({{x} \over {2}}[2.1+(x-1)1])$$
= $$25(49) – {{x( x + 1)} \over {2}}$$………………………………….(ii)
As per the given condition, eq.(i) and eq(ii) are equal to each other;
Therefore,
$${{x(x - 1)} \over {2}} = 25(49) – {{x ( x - 1)} \over {2}}$$
$$x = ±35$$
As we know, the number of house cannot be an a negative number. Hence, the value of x is 35

5. A small terrace at a football ground comprises of 15 steps each of which is 50 m long and built of solid concrete. Each step has a rise of $${{1} \over {4}}$$ m and a tread of $${{1} \over {2}}$$ m. (see Fig. 5.8). Calculate the total volume of concrete required to build the terrace. [Hint : Volume of concrete required to build the first step = $${{1} \over {4}} × {{1} \over {2}} × 50 m^3$$.]

As we can see from the given figure, the first step is $${{1} \over {2}}$$ m wide, 2nd step is 1 m wide and 3rd step is $${{3} \over {2}}$$ m wide. Thus we can understand that the width of step by $${{1} \over {2}}$$ m each time when height is $${{1} \over {4}}$$ m. And also, given length of the steps is 50 m all the time. So, the width of steps forms a series AP in such a way that;
$${{1} \over {2}}$$ , 1, $${{3} \over {2}}$$, 2, ……..
Volume of steps = Volume of Cuboid
Now,
Volume of concrete required to build the first step = $${{1} \over {4}} × {{1} \over {2}} ×50 = {{25} \over {4}}$$
Volume of concrete required to build the second step = $${{1} \over {5}} × {{2} \over {2}} × 50 = {{25} \over {2}}$$
Volume of concrete required to build the second step = $${{1} \over {4}} × {{3} \over {2}}×50 = {{75} \over {2}}$$
Now, we can see the volumes of concrete required to build the steps, are in AP series;
$${{25} \over {4}} , {{25} \over {2}} , {{75} \over {2}}$$…..
Thus, applying the AP series concept,
First term, a = $${{25} \over {4}}$$
Common difference, d = $${{25} \over {2}} – {{25} \over {4}} = {{25} \over {4}}$$
As we know, the sum of n terms is;
$$S_n = {{n} \over {2}}[2a+(n-1)d] = {{15} \over {2}}(2×({{25} \over {4}} )+({{15} \over {2}} -1){{25} \over {4}})$$
Upon solving, we get,
$$S_n = {{15} \over {2}} (100)$$
$$S_n= 750$$
Hence, the total volume of concrete required to build the terrace is $$750 m^3$$.