Khopdi Kharab: Arithmetic Progressions

Exercise 5.1
ANS: 1
(i) It can be observed that
Taxi fare for 1^st km = 15
Taxi fare for first 2 km = 15 + 8 = 23
Taxi fare for first 3 km = 23 + 8 = 31
Taxi fare for first 4 km = 31 + 8 = 39
Clearly 15, 23, 31, 39 … forms an A.P. because every term is 8 more than the preceding term.
(ii) Let the initial volume of air in a cylinder be V lit. In each stroke, the vacuum pump removes

of air remaining in the cylinder at a time. In other words, after every stroke, only

part of air will remain.
Therefore, volumes will be

Clearly, it can be observed that the adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.
(iii) Cost of digging for first metre = 150
Cost of digging for first 2 metres = 150 + 50 = 200
Cost of digging for first 3 metres = 200 + 50 = 250
Cost of digging for first 4 metres = 250 + 50 = 300
Clearly, 150, 200, 250, 300 … forms an A.P. because every term is 50 more than the preceding term.
(iv) We know that if Rs P is deposited at r% compound interest per annum for n years, our money will be

aftern years.
Therefore, after every year, our money will be

Clearly, adjacent terms of this series do not have the same difference between them. Therefore, this is not an A.P.

ANS: 2
(i) a = 10, d= 10
Let the series be a₁, a₂, a₃,a₄, a₅ …
a₁= a = 10
a₂= a₁ + d = 10 + 10 = 20
a₃= a₂ + d = 20 + 10 = 30
a₄= a₃ + d = 30 + 10 = 40
a₅= a₄ + d = 40 + 10 = 50
Therefore, the series will be 10, 20, 30, 40, 50 …
First four terms of this A.P. will be 10, 20, 30, and 40.
(ii) a = −2,d = 0
Let the series be a₁, a₂, a₃,a₄ …
a₁= a = −2
a₂= a₁ + d = − 2 + 0 = −2
a₃= a₂ + d = − 2 + 0 = −2
a₄= a₃ + d = − 2 + 0 = −2
Therefore, the series will be −2, −2, −2, −2 …
First four terms of this A.P. will be −2, −2, −2 and −2.
(iii) a = 4,d = −3
Let the series be a₁, a₂, a₃,a₄ …
a₁= a = 4
a₂= a₁ + d = 4 − 3 = 1
a₃= a₂ + d = 1 − 3 = −2
a₄= a₃ + d = − 2 − 3 = −5
Therefore, the series will be 4, 1, −2 −5 …
First four terms of this A.P. will be 4, 1, −2 and −5.
(iv) a = −1,d =

Let the series be a₁, a₂, a₃,a₄ …

Clearly, the series will be

………….
First four terms of this A.P. will be

.
(v) a = −1.25,d = −0.25
Let the series be a₁, a₂, a₃,a₄ …
a₁= a = −1.25
a₂= a₁ + d = − 1.25 − 0.25 = −1.50
a₃= a₂ + d = − 1.50 − 0.25 = −1.75
a₄= a₃ + d = − 1.75 − 0.25 = −2.00
Clearly, the series will be 1.25, −1.50, −1.75, −2.00 ……..
First four terms of this A.P. will be −1.25, −1.50, −1.75 and −2.00.

ANS: 3

(i) 3, 1, −1, −3…
Here, first term, a = 3
Common difference, d = Second term − First term
= 1 −3 = −2
(ii) −5, −1, 3, 7 …
Here, first term, a = −5
Common difference, d = Second term − First term
= (−1) − (−5) = − 1 + 5 = 4
(iii)

Here, first term,

Common difference, d = Second term − First term

(iv) 0.6, 1.7, 2.8, 3.9…
Here, first term, a = 0.6
Common difference, d = Second term − First term
= 1.7− 0.6
= 1.1

ANS: 4
(i) 2, 4, 8, 16 …
It can be observed that
a₂− a₁ = 4 − 2 = 2
a₃− a₂= 8 − 4 = 4
a₄− a₃ = 16 − 8 = 8
i.e.,a_k₊₁− a_kis not the same every time. Therefore, the given numbers are not forming an A.P.
(ii)

It can be observed that

i.e.,a_k₊₁− a_kis same every time.
Therefore,

and the given numbers are in A.P.
Three more terms are

(iii) −1.2, −3.2,−5.2, −7.2 …
It can be observed that
a₂− a₁ = (−3.2) − (−1.2) = −2
a₃− a₂= (−5.2) − (−3.2) = −2
a₄− a₃ = (−7.2) − (−5.2) = −2
i.e.,a_k₊₁− a_kis same every time. Therefore, d = −2
The given numbers are in A.P.
Three more terms are
a₅= − 7.2 − 2 = −9.2
a₆= − 9.2 − 2 = −11.2
a₇= − 11.2 − 2 = −13.2
(iv) −10, −6,−2, 2 …
It can be observed that
a₂− a₁ = (−6) − (−10) = 4
a₃− a₂= (−2) − (−6) = 4
a₄− a₃ = (2) − (−2) = 4
i.e.,a_k₊₁− a_kis same every time. Therefore, d = 4
The given numbers are in A.P.
Three more terms are
a₅= 2 + 4 = 6
a₆= 6 + 4 = 10
a₇= 10 + 4 = 14
(v)

It can be observed that

i.e.,a_k_{+1 −} a_kis same every time. Therefore,

The given numbers are in A.P.
Three more terms are

(vi) 0.2, 0.22, 0.222, 0.2222 ….
It can be observed that
a₂− a₁ = 0.22 − 0.2 = 0.02
a₃− a₂= 0.222 − 0.22 = 0.002
a₄− a₃ = 0.2222 − 0.222 = 0.0002
i.e.,a_k₊₁− a_kis not the same every time.
Therefore, the given numbers are not in A.P.
(vii) 0, −4, −8,−12 …
It can be observed that
a₂− a₁ = (−4) − 0 = −4
a₃− a₂= (−8) − (−4) = −4
a₄− a₃ = (−12) − (−8) = −4
i.e.,a_k₊₁− a_kis same every time. Therefore, d = −4
The given numbers are in A.P.
Three more terms are
a₅= − 12 − 4 = −16
a₆= − 16 − 4 = −20
a₇= − 20 − 4 = −24
(viii)

It can be observed that

i.e.,a_k₊₁− a_kis same every time. Therefore, d = 0
The given numbers are in A.P.
Three more terms are

(ix) 1, 3, 9, 27 …
It can be observed that
a₂− a₁ = 3 − 1 = 2
a₃− a₂= 9 − 3 = 6
a₄− a₃ = 27 − 9 = 18
i.e.,a_k₊₁− a_kis not the same every time.
Therefore, the given numbers are not in A.P.
(x) a, 2a, 3a, 4a …
It can be observed that
a₂− a₁ = 2a − a = a
a₃− a₂= 3a − 2a = a
a₄− a₃ = 4a − 3a = a
i.e.,a_k₊₁− a_kis same every time. Therefore, d = a
The given numbers are in A.P.
Three more terms are
a₅= 4a + a = 5a
a₆= 5a + a = 6a
a₇= 6a + a = 7a
(xi) a, a²,a³, a⁴ …
It can be observed that
a₂− a₁ = a² − a= a (a − 1)
a₃− a₂= a³ − a²= a² (a − 1)
a₄− a₃ = a⁴ − a³= a³ (a − 1)
i.e.,a_k₊₁− a_kis not the same every time.
Therefore, the given numbers are not in A.P.
(xii)

It can be observed that

i.e.,a_k₊₁− a_kis same every time.
Therefore, the given numbers are in A.P.
And,

Three more terms are

(xiii)

It can be observed that

i.e.,a_k₊₁− a_kis not the same every time.
Therefore, the given numbers are not in A.P.
(xiv) 1², 3², 5², 7² …
Or, 1, 9, 25, 49 …..
It can be observed that
a₂− a₁ = 9 − 1 = 8
a₃− a₂= 25 − 9 = 16
a₄− a₃ = 49 − 25 = 24
i.e.,a_k₊₁− a_kis not the same every time.
Therefore, the given numbers are not in A.P.
(xv) 1², 5², 7², 73 …
Or 1, 25, 49, 73 …
It can be observed that
a₂− a₁ = 25 − 1 = 24
a₃− a₂= 49 − 25 = 24
a₄− a₃ = 73 − 49 = 24
i.e.,a_k₊₁− a_kis same every time.
Therefore, the given numbers are in A.P.
And,d = 24
Three more terms are
a₅= 73+ 24 = 97
a₆= 97 + 24 = 121
a₇= 121 + 24 = 145

Exercise 5.2
ANS: 1
I. a = 7,d = 3, n = 8, a_n = ?
We know that,
For an A.P. a_n = a + (n − 1)d
= 7 + (8 − 1) 3
= 7 + (7) 3
= 7 + 21 = 28
Hence,a_n = 28
II. Given that
a= −18, n = 10, a_n = 0, d= ?
We know that,
a_n= a + (n − 1) d
0 =− 18 + (10 − 1) d
18 = 9d

Hence, common difference, d = 2
III. Given that
d= −3, n = 18, a_n = −5
We know that,
a_n= a + (n − 1) d
−5 = a + (18 − 1) (−3)
−5 = a + (17) (−3)
−5 = a − 51
a= 51 − 5 = 46
Hence,a = 46
IV. a =−18.9, d = 2.5, a_n = 3.6, n= ?
We know that,
a_n= a + (n − 1) d
3.6 = − 18.9 + (n − 1) 2.5
3.6 + 18.9 = (n − 1) 2.5
22.5 = (n − 1) 2.5

Hence,n = 10
V. a = 3.5, d = 0, n = 105, a_n = ?
We know that,
a_n= a + (n − 1) d
a_n= 3.5 + (105 − 1) 0
a_n= 3.5 + 104 × 0
a_n= 3.5
Hence,a_n = 3.5

ANS: 2
I. Given that
A.P. 10, 7, 4, …
First term, a = 10
Common difference, d = a₂ − a₁= 7 − 10
=−3
We know that, a_n = a + (n −1) d
a₃₀= 10 + (30 − 1) (−3)
a₃₀= 10 + (29) (−3)
a₃₀= 10 − 87 = −77
Hence, the correct answer is C.
II. Given that, A.P.

First term a = −3
Common difference, d = a₂ − a₁

We know that,

Hence, the answer is B.

ANS: 3

I.

For this A.P.,
a= 2
a₃= 26
We know that, a_n = a + (n −1) d
a₃= 2 + (3 − 1) d
26 = 2 + 2d
24 = 2d
d= 12
a₂= 2 + (2 − 1) 12
= 14
Therefore, 14 is the missing term.
II.

For this A.P.,
a₂= 13 and
a₄= 3
We know that, a_n = a + (n −1) d
a₂= a + (2 − 1) d
13 = a + d (I)
a₄= a + (4 − 1) d
3 =a + 3d (II)
On subtracting (I) from (II), we obtain
−10 = 2d
d= −5
From equation (I), we obtain
13 = a + (−5)
a= 18
a₃= 18 + (3 − 1) (−5)
= 18 + 2 (−5) = 18 − 10 = 8
Therefore, the missing terms are 18 and 8 respectively.
III.

For this A.P.,

We know that,

Therefore, the missing terms are

and 8 respectively.
IV.

For this A.P.,
a= −4 and
a₆= 6
We know that,
a_n= a + (n − 1) d
a₆= a + (6 − 1) d
6 =− 4 + 5d
10 = 5d
d= 2
a₂= a + d = − 4 + 2 = −2
a₃= a + 2d = − 4 + 2 (2) = 0
a₄= a + 3d = − 4 + 3 (2) = 2
a₅= a + 4d = − 4 + 4 (2) = 4
Therefore, the missing terms are −2, 0, 2, and 4 respectively.
V.

For this A.P.,
a₂= 38
a₆= −22
We know that
a_n= a + (n − 1) d
a₂= a + (2 − 1) d
38 = a + d (1)
a₆= a + (6 − 1) d
−22 = a + 5d (2)
On subtracting equation (1) from (2), we obtain
−22 − 38 = 4d
−60 = 4d
d= −15
a= a₂ − d = 38 − (−15) = 53
a₃= a + 2d = 53 + 2 (−15) = 23
a₄= a + 3d = 53 + 3 (−15) = 8
a₅= a + 4d = 53 + 4 (−15) = −7
Therefore, the missing terms are 53, 23, 8, and −7 respectively.

ANS: 4
3, 8, 13, 18, …
For this A.P.,
a = 3
d = a₂− a₁ = 8 − 3 = 5
Let n^thterm of this A.P. be 78.
a_n= a + (n − 1) d
78 = 3 + (n −1) 5
75 = (n −1) 5
(n − 1) = 15
n = 16
Hence, 16^thterm of this A.P. is 78.

ANS: 5
I. 7, 13, 19, …,205
For this A.P.,
a= 7
d= a₂ − a₁ = 13 − 7 = 6
Let there are n terms in this A.P.
a_n= 205
We know that
a_n= a + (n − 1) d
Therefore, 205 = 7 + (n − 1) 6
198 = (n − 1) 6
33 = (n − 1)
n= 34
Therefore, this given series has 34 terms in it.
II.

For this A.P.,

Let there are n terms in this A.P.
Therefore,a_n = −47 and we know that,

Therefore, this given A.P. has 27 terms in it.

ANS: 6
For this A.P.,
a = 11
d = a₂− a₁ = 8 − 11 = −3
Let −150 be then^th term of this A.P.
We know that,

Clearly, n is not an integer.
Therefore, −150 is not a term of this A.P.

ANS: 7

Given that,
a₁₁ = 38
a₁₆ = 73
We know that,
a_n= a + (n − 1) d
a₁₁ =a + (11 − 1) d
38 = a + 10d (1)
Similarly,
a₁₆ =a + (16 − 1) d
73 = a + 15d (2)
On subtracting (1) from (2), we obtain
35 = 5d
d = 7
From equation (1),
38 = a + 10 × (7)
38 − 70 = a
a = −32
a₃₁ =a + (31 − 1) d
=− 32 + 30 (7)
=− 32 + 210
= 178
Hence, 31^stterm is 178.

ANS: 8
Given that,
a₃ = 12
a₅₀ = 106
We know that,
a_n= a + (n − 1) d
a₃ =a + (3 − 1) d
12 = a + 2d (I)
Similarly, a₅₀= a + (50 − 1) d
106 = a + 49d (II)
On subtracting (I) from (II), we obtain
94 = 47d
d = 2
From equation (I), we obtain
12 = a + 2 (2)
a = 12 − 4 = 8
a₂₉ =a + (29 − 1) d
a₂₉ = 8 + (28)2
a₂₉ = 8 + 56 = 64
Therefore, 29^thterm is 64.

ANS: 9
Given that,
a₃ = 4
a₉ = −8
We know that,
a_n= a + (n − 1) d
a₃ = a + (3 − 1) d
4 = a + 2d (I)
a₉ =a + (9 − 1) d
−8 = a + 8d (II)
On subtracting equation (I) from (II), we obtain
−12 = 6d
d = −2
From equation (I), we obtain
4 = a + 2 (−2)
4 = a − 4
a = 8
Let n^thterm of this A.P. be zero.
a_n=a + (n − 1) d
0 = 8 + (n −1) (−2)
0 = 8 − 2n+ 2
2n = 10
n = 5
Hence, 5^thterm of this A.P. is 0.

ANS: 10

We know that,
For an A.P., a_n= a + (n − 1) d
a₁₇ =a + (17 − 1) d
a₁₇ =a + 16d
Similarly, a₁₀= a + 9d
It is given that
a₁₇ −a₁₀ = 7
(a + 16d)− (a + 9d) = 7
7d = 7
d = 1
Therefore, the common difference is 1.

ANS: 11
Given A.P. is 3, 15, 27, 39, …
a = 3
d = a₂− a₁ = 15 − 3 = 12
a₅₄ =a + (54 − 1) d
= 3 + (53) (12)
= 3 + 636 = 639
132 + 639 = 771
We have to find the term of this A.P. which is 771.
Let n^thterm be 771.
a_n= a + (n − 1) d
771 = 3 + (n −1) 12
768 = (n −1) 12
(n − 1) = 64
n = 65
Therefore, 65^thterm was 132 more than 54^th term.
Alternatively,
Let n^thterm be 132 more than 54^th term.

ANS: 12
Let the first term of these A.P.s be a₁ and a₂respectively and the common difference of these A.P.s be d.
For first A.P.,
a₁₀₀= a₁ + (100 − 1) d
=a₁ + 99d
a₁₀₀₀= a₁ + (1000 − 1) d
a₁₀₀₀= a₁ + 999d
For second A.P.,
a₁₀₀= a₂ + (100 − 1) d
=a₂ + 99d
a₁₀₀₀= a₂ + (1000 − 1) d
=a₂ + 999d
Given that, difference between
100^th term of these A.P.s = 100
Therefore, (a₁+ 99d) − (a₂ + 99d) = 100
a₁ −a₂ = 100 (1)
Difference between 1000^th terms of these A.P.s
(a₁ + 999d) − (a₂ + 999d) = a₁− a₂
From equation (1),
This difference, a₁− a₂= 100
Hence, the difference between 1000^th terms of these A.P. will be 100.

ANS: 13
First three-digit number that is divisible by 7 = 105
Next number = 105 + 7 = 112
Therefore, 105, 112, 119, …
All are three digit numbers which are divisible by 7 and thus, all these are terms of an A.P. having first term as 105 and common difference as 7.
The maximum possible three-digit number is 999. When we divide it by 7, the remainder will be 5. Clearly, 999 − 5 = 994 is the maximum possible three-digit number that is divisible by 7.
The series is as follows.
105, 112, 119, …,994
Let 994 be the nth term of this A.P.
a = 105
d = 7
a_n= 994
n = ?
a_n= a + (n − 1) d
994 = 105 + (n −1) 7
889 = (n −1) 7
(n − 1) = 127
n = 128
Therefore, 128 three-digit numbers are divisible by 7.

ANS: 14
First multiple of 4 that is greater than 10 is 12. Next will be 16.
Therefore, 12, 16, 20, 24, …
All these are divisible by 4 and thus, all these are terms of an A.P. with first term as 12 and common difference as 4.
When we divide 250 by 4, the remainder will be 2. Therefore, 250 − 2 = 248 is divisible by 4.
The series is as follows.
12, 16, 20, 24, …,248
Let 248 be the n^thterm of this A.P.

Therefore, there are 60 multiples of 4 between 10 and 250.

ANS: 15
63, 65, 67, …
a = 63
d = a₂ − a₁ = 65 −63 = 2
n^thterm of this A.P. = a_n = a + (n− 1) d
a_n= 63 + (n − 1) 2 = 63 + 2n − 2
a_n = 61 + 2n (1)
3, 10, 17, …
a = 3
d = a₂− a₁ = 10 − 3 = 7
n^thterm of this A.P. = 3 + (n − 1) 7
a_n= 3 + 7n − 7
a_n= 7n − 4 (2)
It is given that, n^thterm of these A.P.s are equal to each other.
Equating both these equations, we obtain
61 + 2n = 7n− 4
61 + 4 = 5n
5n = 65
n = 13
Therefore, 13^thterms of both these A.P.s are equal to each other.

ANS: 16

=a₃ = 16
a + (3 −1) d = 16
a + 2d = 16 (1)
a₇ −a₅ = 12
[a+ (7 −1) d] − [a + (5 − 1) d]= 12
(a + 6d)− (a + 4d) = 12
2d = 12
d = 6
From equation (1), we obtain
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be
4, 10, 16, 22, …

ANS: 17
Given A.P. is
3, 8, 13, …, 253
Common difference for this A.P. is 5.
Therefore, this A.P. can be written in reverse order as
253, 248, 243, …,13, 8, 5
For this A.P.,
a = 253
d = 248 −253 = −5
n = 20
a₂₀ =a + (20 − 1) d
a₂₀ = 253 + (19) (−5)
a₂₀ = 253 − 95
a = 158
Therefore, 20^thterm from the last term is 158.

ANS: 18
We know that,
a_n = a + (n − 1) d
a₄ = a + (4 − 1) d
a₄ = a + 3d
Similarly,
a₈ = a + 7d
a₆ = a + 5d
a₁₀ = a + 9d
Given that, a₄ + a₈ = 24
a + 3d + a + 7d = 24
2a + 10d = 24
a + 5d = 12 (1)
a₆ + a₁₀ = 44
a + 5d + a + 9d = 44
2a + 14d = 44
a + 7d = 22 (2)
On subtracting equation (1) from (2), we obtain
2d = 22 − 12
2d = 10
d = 5
From equation (1), we obtain
a + 5d = 12
a + 5 (5) = 12
a + 25 = 12
a = −13
a₂ = a + d = − 13 + 5 = −8
a₃ = a₂ + d = − 8 + 5 = −3
Therefore, the first three terms of this A.P. are −13, −8, and −3.

ANS: 19
It can be observed that the incomes that Subba Rao obtained in various years are in A.P. as every year, his salary is increased by Rs 200.
Therefore, the salaries of each year after 1995 are
5000, 5200, 5400, …
Here, a = 5000
d = 200
Let after n^thyear, his salary be Rs 7000.
Therefore, a_n= a + (n − 1) d
7000 = 5000 + (n− 1) 200
200(n − 1) = 2000
(n − 1) = 10
n = 11
Therefore, in 11th year, his salary will be Rs 7000.

ANS: 20
Given that,
a = 5
d = 1.75
a_n= 20.75
n = ?
a_n= a + (n − 1) d

n − 1 = 9
n = 10
Hence, n is 10.

Exercise 5.3
ANS: 1
(i)2, 7, 12 ,…,to 10 terms
For this A.P.,
a= 2
d= a₂ − a₁ = 7 − 2 = 5
n= 10
We know that,

(ii)−37,−33, −29 ,…, to 12 terms
For this A.P.,
a= −37
d= a₂ − a₁ = (−33) −(−37)
=− 33 + 37 = 4
n= 12
We know that,

(iii) 0.6, 1.7, 2.8 ,…,to 100 terms
For this A.P.,
a= 0.6
d= a₂ − a₁ = 1.7 − 0.6 = 1.1
n= 100
We know that,

(iv)

.…….. , to 11 terms
For this A.P.,

n= 11

We know that,

ANS: 2
(i)7 +

+ 14 + …………+ 84
For this A.P.,
a= 7
l= 84

Let 84 be the n^thterm of this A.P.
l= a + (n − 1)d

22 = n − 1
n= 23
We know that,

(ii)34 + 32 + 30 + ……….. + 10
For this A.P.,
a= 34
d= a₂ − a₁ = 32 − 34 =−2
l= 10
Let 10 be the n^th term of this A.P.
l= a + (n − 1) d
10 = 34 + (n − 1) (−2)
−24 = (n − 1) (−2)
12 = n − 1
n= 13

(iii)(−5) + (−8) + (−11) + ………… + (−230)
For this A.P.,
a= −5
l= −230
d= a₂ − a₁ = (−8) −(−5)
= −8 + 5 = −3
Let−230 be the n^th term of this A.P.
l= a + (n − 1)d
−230 = − 5 + (n − 1) (−3)
−225 = (n − 1) (−3)
(n− 1) = 75
n= 76
And,

ANS: 3
(i) Given that, a= 5, d = 3, a_n = 50
Asa_n = a + (n − 1)d,
∴ 50 = 5 + (n − 1)3
45 = (n − 1)3
15 = n − 1
n= 16

(ii) Given that, a= 7, a₁₃ = 35
Asa_n = a + (n − 1) d,
∴a₁₃ = a + (13 − 1) d
35 = 7 + 12 d
35− 7 = 12d
28 = 12d

(iii)Given that, a₁₂ = 37, d = 3
Asa_n = a + (n − 1)d,
a₁₂= a + (12 − 1)3
37 = a + 33
a= 4

(iv) Given that, a₃= 15, S₁₀ = 125
Asa_n = a + (n − 1)d,
a₃= a + (3 − 1)d
15 = a + 2d (i)

On multiplying equation (1) by 2, we obtain
30 = 2a + 4d (iii)
On subtracting equation (iii) from (ii), we obtain
−5 = 5d
d= −1
From equation (i),
15 = a + 2(−1)
15 = a − 2
a= 17
a₁₀= a + (10 − 1)d
a₁₀= 17 + (9) (−1)
a₁₀= 17 − 9 = 8
(v)Given that,d = 5, S₉ = 75
As

25 = 3(a + 20)
25 = 3a + 60
3a= 25 − 60

a_n= a + (n − 1)d
a₉= a + (9 − 1) (5)

(vi) Given that, a= 2, d = 8, S_n = 90
As

90 = n [2 + (n − 1)4]
90 = n [2 + 4n − 4]
90 = n (4n − 2) = 4n² − 2n
4n²− 2n − 90 = 0
4n²− 20n + 18n − 90 = 0
4n(n − 5) + 18 (n − 5) = 0
(n− 5) (4n + 18) = 0
Eithern − 5 = 0 or 4n + 18 = 0
n= 5 or

However,n can neither be negative nor fractional.
Therefore,n = 5
a_n= a + (n − 1)d
a₅= 2 + (5 − 1)8
= 2 + (4) (8)
= 2 + 32 = 34
(vii) Given that, a= 8, a_n = 62, S_n = 210

n= 6
a_n= a + (n − 1)d
62 = 8 + (6 − 1)d
62− 8 = 5d
54 = 5d

(viii) Given that, a_n= 4, d = 2, S_n = −14
a_n= a + (n − 1)d
4 =a + (n − 1)2
4 =a + 2n − 2
a+ 2n = 6
a= 6 − 2n (i)

−28 = n (a + 4)
−28 = n (6 − 2n + 4) {From equation (i)}
−28 = n (− 2n + 10)
−28 = − 2n² + 10n
2n²− 10n − 28 = 0
n²− 5n −14 = 0
n²− 7n + 2n − 14 = 0
n(n − 7) + 2(n − 7) = 0
(n− 7) (n + 2) = 0
Eithern − 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 obtain
a= 6 − 2n
a= 6 − 2(7)
= 6 − 14
=−8
(ix)Given that,a = 3, n = 8, S = 192

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.

(16) × (2) = a + 28
32 = a + 28
a= 4

ANS: 4
Let there be nterms of this A.P.
For this A.P., a= 9
d = a₂− a₁ = 17 − 9 = 8

636 = n [9 + 4n− 4]
636 = n (4n+ 5)
4n² + 5n − 636 = 0
4n² + 53n − 48n − 636 = 0
n (4n + 53) − 12 (4n + 53) = 0
(4n + 53) (n− 12) = 0
Either 4n + 53 = 0 or n − 12 = 0

or n = 12
n cannot be

. As the number of terms can neither be negative nor fractional, therefore, n = 12 only.

ANS: 5
Given that,
a = 5
l = 45
S_n= 400

n = 16
l = a + (n− 1) d
45 = 5 + (16 − 1)d
40 = 15d

ANS: 6
Given that,
a = 17
l = 350
d = 9
Let there be nterms in the A.P.
l = a + (n− 1) d
350 = 17 + (n −1)9
333 = (n −1)9
(n − 1) = 37
n = 38

Thus, this A.P. contains 38 terms and the sum of the terms of this A.P. is 6973.

ANS: 7
d = 7
a₂₂ = 149
S₂₂ = ?
a_n= a + (n − 1)d
a₂₂ =a + (22 − 1)d
149 = a + 21 × 7
149 = a + 147
a = 2

ANS: 8
Given that,
a₂ = 14
a₃ = 18
d = a₃− a₂ = 18 − 14 = 4
a₂ =a + d
14 = a + 4
a = 10

= 5610

ANS: 9
Given that,
S₇ = 49
S₁₇ = 289

7 = (a + 3d)
a + 3d = 7 (i)
Similarly,

17 = (a + 8d)
a + 8d = 17 (ii)
Subtracting equation (i) from equation (ii),
5d = 10
d = 2
From equation (i),
a + 3(2) = 7
a + 6 = 7
a = 1

= n²

^{ANS: 10}
^{(i) a_n= 3 + 4n

a₁= 3 + 4(1) = 7

a₂= 3 + 4(2) = 3 + 8 = 11

a₃= 3 + 4(3) = 3 + 12 = 15

a₄= 3 + 4(4) = 3 + 16 = 19

It can be observed that

a₂− a₁ = 11 − 7 = 4

a₃− a₂ = 15 − 11 = 4

a₄− a₃ = 19 − 15 = 4

i.e.,a_k_{+ 1} − a_kis same every time. Therefore, this is an AP with common difference as 4 and first term as 7.

= 15 × 35

= 525

(ii) a_n= 9 − 5n

a₁= 9 − 5 × 1 = 9 − 5 = 4

a₂= 9 − 5 × 2 = 9 − 10 = −1

a₃= 9 − 5 × 3 = 9 − 15 = −6

a₄= 9 − 5 × 4 = 9 − 20 = −11

It can be observed that

a₂− a₁ = − 1 − 4 = −5

a₃− a₂ = − 6 − (−1) = −5

a₄− a₃ = − 11 − (−6) = −5

i.e.,a_k_{+ 1} − a_kis same every time. Therefore, this is an A.P. with common difference as −5 and first term as 4.

=−465

ANS: 11}

Given that,
S_n= 4n − n²
First term, a =S₁ = 4(1) − (1)² = 4 − 1 = 3
Sum of first two terms = S₂
= 4(2) − (2)²= 8 − 4 = 4
Second term, a₂= S₂ − S₁ = 4 − 3 = 1
d = a₂− a = 1 − 3 = −2
a_n= a + (n − 1)d
= 3 + (n −1) (−2)
= 3 − 2n + 2
= 5 − 2n
Therefore, a₃= 5 − 2(3) = 5 − 6 = −1
a₁₀ = 5 − 2(10) = 5 − 20 = −15
Hence, the sum of first two terms is 4. The second term is 1. 3^rd, 10^th, and n^th terms are −1, −15, and 5 −2n respectively.

ANS: 12

The positive integers that are divisible by 6 are
6, 12, 18, 24 …
It can be observed that these are making an A.P. whose first term is 6 and common difference is 6.
a = 6
d = 6
S₄₀=?

= 20[12 + (39) (6)]
= 20(12 + 234)
= 20 × 246
= 4920

ANS: 13
The multiples of 8 are
8, 16, 24, 32…
These are in an A.P., having first term as 8 and common difference as 8.
Therefore, a = 8
d = 8
S₁₅=?

= 960

ANS: 14
The odd numbers between 0 and 50 are
1, 3, 5, 7, 9 …49
Therefore, it can be observed that these odd numbers are in an A.P.
a = 1
d = 2
l = 49
l = a + (n − 1) d
49 = 1 + (n −1)2
48 = 2(n −1)
n − 1 = 24
n = 25

= 625

ANS: 15
It can be observed that these penalties are in an A.P. having first term as 200 and common difference as 50.
a = 200
d = 50
Penalty that has to be paid if he has delayed the work by 30 days = S₃₀

= 15 [400 + 1450]
= 15 (1850)
= 27750
Therefore, the contractor has to pay Rs 27750 as penalty.

ANS: 16
Let the cost of 1^st prize be P.
Cost of 2^ndprize = P − 20
And cost of 3^rdprize = P − 40
It can be observed that the cost of these prizes are in an A.P. having common difference as−20 and first term as P.
a = P
d = −20
Given that, S₇= 700

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.

ANS: 17
It 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

= 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.

ANS: 18
Semi-perimeter of circle = πr
I₁ = π(0.5)

I₂ = π(1) = πcm
I₃ = π(1.5) =

Therefore, I₁, I₂, I₃ ,i.e. the lengths of the semi-circles are in an A.P.,

S₁₃=?
We know that the sum ofn terms of an a A.P. is given by

= 143
Therefore, the length of such spiral of thirteen consecutive semi-circles will be 143 cm.

ANS: 19
It can be observed that the numbers of logs in rows are in an A.P.
20, 19, 18…
For this A.P.,
a = 20
d = a₂− a₁ = 19 − 20 = −1
Let a total of 200 logs be placed in n rows.
S_n= 200

400 = n (40 −n + 1)
400 = n (41 −n)
400 = 41n −n²
n² −41n + 400 = 0
n² −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
a_n= a + (n − 1)d
a₁₆ = 20 + (16 − 1) (−1)
a₁₆ = 20 − 15
a₁₆ = 5
Similarly,
a₂₅ = 20 + (25 − 1) (−1)
a₂₅ = 20 − 24
= −4
Clearly, the number of logs in 16^th row is 5. However, the number of logs in 25^throw is negative, which is not possible.
Therefore, 200 logs can be placed in 16 rows and the number of logs in the 16^throw is 5.

ANS: 20

The distances of potatoes are as follows.
5, 8, 11, 14…
It can be observed that these distances are in A.P.
a = 5
d = 8 − 5 = 3

= 5[10 + 9 × 3]
= 5(10 + 27) = 5(37)
= 185
As every time she has to run back to the bucket, therefore, the total distance that the competitor has to run will be two times of it.
Therefore, total distance that the competitor will run = 2 × 185
= 370 m
Alternatively,
The distances of potatoes from the bucket are 5, 8, 11, 14…
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 are
10, 16, 22, 28, 34,……….
a = 10
d = 16 −10 = 6
S₁₀=?

= 5[20 + 54]
= 5 (74)
= 370
Therefore, the competitor will run a total distance of 370 m.

Exercise 5.4
ANS: 1
Given A.P. is 121, 117, 113 …
a = 121
d = 117 −121 = −4
a_n= a + (n − 1) d
= 121 + (n − 1) (−4)
= 121 − 4n + 4
= 125 − 4n
We have to find the first negative term of this A.P.

Therefore, 32^ndterm will be the first negative term of this A.P.

ANS: 2
We know that,
a_n= a + (n − 1) d
a₃ = a + (3 − 1) d
a₃ = a + 2d
Similarly, a₇ = a + 6d
Given that, a₃ + a₇ = 6
(a + 2d) + (a + 6d) = 6
2a + 8d = 6
a + 4d = 3
a = 3 − 4d (i)
Also, it is given that (a₃) × (a₇) = 8
(a + 2d) × (a+ 6d) = 8
From equation (i),

From equation (i),

ANS: 3
It is given that the rungs are 25 cm apart and the top and bottom rungs are

m apart.
∴ Total number of rungs

Now, as the lengths of the rungs decrease uniformly, they will be in an A.P.
The length of the wood required for the rungs equals the sum of all the terms of this A.P.
First term, a = 45
Last term, l = 25
n= 11

Therefore, the length of the wood required for the rungs is 385 cm.

ANS: 4
The number of houses was
1, 2, 3 … 49
It can be observed that the number of houses are in an A.P. having a as 1 and dalso as 1.
Let us assume that the number of x^thhouse was like this.
We know that,
Sum of n terms in an A.P.

Sum of number of houses preceding x^th house = S_x_{− 1}

Sum of number of houses following x^th house = S₄₉ −S_x

It is given that these sums are equal to each other.

However, the house numbers are positive integers.
The value of xwill be 35 only.
Therefore, house number 35 is such that the sum of the numbers of houses preceding the house numbered 35 is equal to the sum of the numbers of the houses following it.

ANS: 5