# Class 10 RD SHARMA Solutions Maths Chapter 5 - Arithmetic Progressions

## Arithmetic Progressions Exercise Ex. 5.1

### Solution 1 (i)

### Solution 1 (ii)

### Solution 1 (iii)

### Solution 1 (iv)

### Solution 1 (v)

### Solution 1 (vi)

### Solution 1 (vii)

### Solution 1 (viii)

### Solution 1 (ix)

### Solution 2 (i)

### Solution 2 (ii)

### Solution 2 (iii)

### Solution 2 (iv)

### Solution 2 (v)

### Solution 3 (i)

### Solution 3 (ii)

### Solution 3 (iii)

### Solution 3 (iv)

## Arithmetic Progressions Exercise Ex. 5.2

### Solution 1

### Solution 2

### Solution 3

### Solution 4 (i)

### Solution 4 (ii)

### Solution 4 (iii)

### Solution 4 (iv)

### Solution 5

### Solution 6(i)

### Solution 6(ii)

### Solution 6(iii)

## Arithmetic Progressions Exercise Ex. 5.3

### Solution 1

### Solution 2

### Solution 3 (i)

### Solution 3 (ii)

### Solution 3(iii)

### Solution 4 (i)

### Solution 4 (ii)

### Solution 4 (iii)

### Solution 4 (iv)

### Solution 5 (i)

### Solution 5 (ii)

### Solution 5 (iii)

### Solution 5 (iv)

### Solution 5 (v)

### Solution 5 (vi)

### Solution 5 (vii)

### Solution 5 (viii)

### Solution 5 (ix)

### Solution 5 (x)

### Solution 5 (xi)

### Solution 5 (xii)

### Solution 6

### Solution 7

## Arithmetic Progressions Exercise Ex. 5.4

### Solution 1 (i)

### Solution 1 (ii)

### Solution 1 (iii)

### Solution 1 (iv)

### Solution 1 (v)

### Solution 1 (vi)

### Solution 1 (vii)

### Solution 2 (i)

### Solution 2 (ii)

### Solution 2 (iii)

### Solution 2 (iv)

### Solution 2 (v)

### Solution 2 (vi)

The given A.P. is -7, -12, -17, -22,…

First term (a) = -7

Common difference = -12 - (-7) = -5

Suppose nth term of the A.P. is -82.

So, -82 is the 16^{th} term of the A.P.

To check whether -100
is any term of the A.P., take a_{n} as -100.

So, n is not a natural number.

Hence, -100 is not the term of this A.P.

### Solution 3 (i)

### Solution 3 (ii)

### Solution 3 (iii)

### Solution 4 (i)

### Solution 4 (ii)

### Solution 4 (iii)

### Solution 4 (iv)

### Solution 5

### Solution 6

### Solution 7

### Solution 8

### Solution 9 (i)

### Solution 9 (ii)

A.P. is 3, 8, 13, ..., 253

We have:

Last term (*l*) = 253

Common difference (*d*) = 8 - 3 = 5

Therefore,

12^{th} term from end

= *l* - (*n* - 1)*d*

= 253 - (12 - 1) (5)

= 253 - 55

= 198

### Solution 9 (iii)

### Solution 10

### Solution 11

### Solution 12

### Solution 13

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 20 (i)

### Solution 20 (ii)

### Solution 21

### Solution 22 (i)

### Solution 22 (ii)

### Solution 22 (iii)

### Solution 22 (iv)

### Solution 23

### Solution 24

### Solution 25

### Solution 26

### Solution 27

### Solution 28

### Solution 29

### Solution 30

### Solution 31

### Solution 32

### Solution 33

### Solution 34

### Solution 35

### Solution 36

### Solution 37

Thus, n^{th} term is given by

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

a_{n} = 3 + (n - 1)4

a_{n} = 3 + 4n - 4

a_{n} = 4n - 1

### Solution 38

The smallest three digit number divisible by 9 = 108

The largest three digit number divisible by 9 = 999

Here let us write the series in this form,

108, 117, 126, …………….., 999

a = 108, d = 9

t_{n} = a + (n - 1)d

999= 108 + (n - 1)9

⇒ 999 - 108 = (n - 1)9

⇒ 891 = (n - 1)9

⇒ (n - 1) = 99

⇒ n = 99 + 1

∴ n = 100

Number of terms divisible by 9

Number of all three digit natural numbers divisible by 9 is 100.

### Solution 39

### Solution 40

### Solution 41

Let the first term be 'a' and the common difference be 'd'

t_{24} = a + (24 - 1)d = a + 23d

t_{10} = a + (10 - 1)d = a + 9d

t_{72} = a + (72 - 1)d = a + 71d

t_{15} = a + (15 - 1)d = a + 14d

t_{24} = 2t_{10}

⇒ a + 23d = 2(a + 9d)

⇒ a + 23d = 2a + 18d

⇒ 23d - 18d = 2a - a

∴ 5d = a

t_{72} = a + 71d

= 5d + 71d

= 76d

= 20d + 56d

= 4 × 5d + 4 × 14d

= 4(5d + 14d)

= 4(a + 14d)

= 4t_{15}

∴t_{72} = 4t_{15}

### Solution 42

L.C.M. of 2 and 5 = 10

3- digit number after 100 divisible by 10 = 110

3- digit number before 999 divisible by 10 = 990

Let the number of natural numbers be 'n'

990 = 110 + (n - 1)d

⇒ 990 - 110 = (n - 1) × 10

⇒ 880 = 10 × (n - 1)

⇒ n - 1 = 88

∴ n = 89

The number of natural numbers between 110 and 999 which are divisible by 2 and 5 is 89.

### Solution 43

Let the first term be 'a' and the common difference be 'd'.

### Solution 44

### Solution 45

Let the first term be 'a' and the common difference be 'd'.

a = 40

d = 37 - 40 = - 3

Let the n^{th} term of the series be 0.

t_{n} = a + (n - 1)d

⇒ 0 = 40 + (n - 1)( - 3)

⇒ 0 = 40 - 3(n - 1)

⇒ 3(n - 1) = 40

∴ No term of the series is 0.

### Solution 46

Given A.P. is 213, 205, 197, …, 37.

Here, first term = a = 213

And, common difference = d = 205 - 213 = -8

a_{n} = 37

n^{th} term of an A.P. is given by

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

⇒ 37 = 213 + (n - 1)(-8)

⇒ 37 = 213 - 8n + 8

⇒ 37 = 221 - 8n

⇒ 8n = 221 - 37

⇒ 8n = 184

⇒ n = 23

So, there are 23 terms in the given A.P.

⇒ The middle term is 12^{th }term.

⇒ a_{12} = 213 + (12 - 1)(-8)

= 213 + (11)(-8)

= 213 - 88

= 125

Hence, the middle term is 125.

### Solution 47

Let a be the first term and d be the common difference of the A.P.

Then, we have

a_{5} = 31 and a_{25} = a_{5} + 140

⇒ a + 4d = 31 and a + 24d = a + 4d + 140

⇒ a + 4d = 31 and 20d = 140

⇒ a + 4d = 31 and d = 7

⇒ a + 4(7) = 31 and d = 7

⇒ a + 28 = 31 and d = 7

⇒ a = 3 and d = 7

Hence, the A.P. is a, a + d, a + 2d, a + 3d, ……

i.e. 3, 3 + 7, 3 + 2(7), 3 + 3(7), ……

i.e. 3, 10, 17, 24, …..

### Solution 48

### Solution 49

### Solution 50

### Solution 51

### Solution 52

### Solution 53

### Solution 54

## Arithmetic Progressions Exercise Ex. 5.5

### Solution 1

### Solution 2

### Solution 3

### Solution 5

### Solution 4

### Solution 6

### Solution 7

### Solution 8

### Solution 9

Let the first three terms of an A.P. be a - d, a, a + d

As per the question,

(a - d) + a + (a + d) = 18

∴ 3a = 18

∴ a = 6

Also, (a - d)(a + d) = 5d

∴ (6 - d)(6 + d) = 5d

∴ 36
- d^{2} = 5d

∴ d^{2}
+ 5d - 36 = 0

∴ d^{2}
+ 9d - 4d - 36 = 0

∴ (d + 9)(d - 4) = 0

∴ d = -9 or d = 4

Thus, the terms will be 15, 6, -3 or 2, 6, 10.

### Solution 10

### Solution 11

### Solution 12

## Arithmetic Progressions Exercise Ex. 5.6

### Solution 1 (i)

### Solution 1 (ii)

### Solution 1 (iii)

### Solution 1 (iv)

### Solution 1 (v)

### Solution 1 (vi)

### Solution 1 (vii)

### Solution 1 (viii)

### Solution 2

### Solution 3

### Solution 4

### Solution 5 (i)

### Solution 5 (ii)

### Solution 5 (iii)

### Solution 5 (iv)

### Solution 6

### Solution 7

### Solution 8

### Solution 9

### Solution 10 (i)

### Solution 10 (ii)

### Solution 10 (iii)

### Solution 10 (iv)

### Solution 10(v)

### Solution 11 (i)

### Solution 10 (vi)

Let the required number of terms be n.

As the given A.P. is 45, 39, 33 …

Here, a = 45 and d = 39 - 45 = -6

The sum is given as 180

∴ S_{n} = 180

When n = 10,

When n = 6,

Hence, number of terms can be 6 or 10.

### Solution 11 (ii)

### Solution 11 (iii)

### Solution 12(i)

Multiples of 8 are8,16,24,…

Now,

n=15, a=8, d=8

### Solution 12(ii)

a) divisible by 3 3,6,9,… Now, n=40, a=3, d=3

b) divisible by 5 5,10,15,… Now, n=40, a=5, d=5

c)divisible by 6 6,12,18,… Now, n=40, a=3, d=6

### Solution 12(iii)

Three-digit numbers divisible by 13 are 104,117,
130,…988.
Now,
a=104, *l*=988

### Solution 12(iv)

Three-digit numbers which are multiples
of 11 are 110,121, 132,…990.
Now,
a=110, *l*=990

### Solution 12(v)

Two-digit numbers divisible by 4 are 12,16,…96.
Now,
a=12, *l*=96

### Solution 12 (vi)

The multiples of 3 are 3, 6, 9, 12, 15, 18, 21, …

These are in A.P. with,

first term (a) = 3 and common difference (d) = 3

To find S_{8} when a = 3, d = 3

Hence, the sum of first 8 multiples of 3 is 108.

### Solution 13 (i)

### Solution 13 (ii)

### Solution 13 (iii)

### Solution 13 (iv)

### Solution 13 (v)

### Solution 13 (vi)

### Solution 13 (vii)

### Solution 13 (viii)

Let 'a' be the first term and 'd' be the common difference.

t_{n} = a + (n - 1)d

### Solution 14

### Solution 15

### Solution 16

### Solution 17

### Solution 18

### Solution 19

### Solution 20

### Solution 21

### Solution 22 (i)

### Solution 22 (ii)

Sum of first n terms of an AP is given by

As per the question, S_{4} = 40 and S_{14}
= 280

Also,

Subtracting (i) from (ii), we get, 10d = 20

Therefore, d = 2

Substituting d in (i), we get, a = 7

Sum of first n terms becomes

### Solution 23

### Solution 24

### Solution 25

### Solution 26

Let the number of terms be 'n', 'a' be the first term and 'd' be the common difference.

t_{n} = a + (n - 1)d

⇒ 49 = 7 + (n - 1)d

⇒ 42 = (n - 1)d…..(i)

∴ 840 = n[14 + (n - 1)d]……(ii)

Substituting (ii) in (i),

840 = n[14 + 42]

⇒ 840 = 56n

∴ n = 15

Substituting n in (i)

42 = (15 - 1)d

Common difference, d =3

### Solution 27

Let the number of terms be 'n', 'a' be the first term and 'd' be the common difference.

t_{n} = a + (n - 1)d

⇒ 45 = 5 + (n - 1)d

⇒ 40 = (n - 1)d…..(i)

∴ 800 = n[10 + (n - 1)d]……(ii)

Substituting (ii) in (i),

800 = n[10 + 40]

⇒ 800 = 50n

∴ n = 16

Substituting n in (i)

40 = (16 - 1)d

### Solution 28

### Solution 29

Let 'a' be the first term and 'd' be the common difference.

t_{n} = a + (n - 1)d

t_{10} = a + (10 - 1)d

⇒ 21 = a + 9d……(i)

120 = 5[2a + 9d]

24 = 2a + 9d………(ii)

(ii) - (i) ⇒

a = 3

Substituting a in (i), we get

a + 9d = 21

⇒ 3 + 9d = 21

⇒ 9d = 18

∴d = 2

t_{n} = a + (n - 1)d

= 3 + (n - 1)2

= 3 + 2n - 2

∴ t_{n}= 2n + 1

### Solution 30

Let the first term be 'a' and the common difference be 'd'

63 = 7[a + 3d]

9 = a + 3d……….(i)

Sum of the next 7 terms = 161

Sum of the first 14 terms = 63 + 161 = 224

224 = 7[2a + 13d]

32 = 2a + 13d………..(ii)

Solving (i) and (ii), we get

d = 2, a = 3

28^{th} term of the A.P., t_{28} = a + (28 - 1)d

= 3 + 27 × 2

= 3 + 54

= 57

∴ The 28^{th} of the A.P. is 57.

### Solution 31

Let the first term be 'a' and the common difference be 'd'.

26 × 2 = [2a + (7 - 1)d]

52 = 2a + 6d

26 = a + 3d……..(i)

From (i) and (ii),

⇒ 13d = 104

∴d = 8

From (i), a = 2

The A.P. is 2, 10, 18, 26,…….

### Solution 32

Let the first term of the A.P. be 'a' and the common difference be 'd'.

t_{n} = - 4n + 15

t_{1} = - 4 × 1 + 15 = 11

t_{2} = - 4 × 2 + 15 = 7

t_{3} = - 4 × 3 + 15 = 3

Common Difference, d = 7 - 11 = -4

= 10 × (-54)

= - 540

*Note: Answer given in the book is incorrect.

### Solution 33

### Solution 34

### Solution 35

### Solution 36

### Solution 37

Let the number of the terms be 'n'.

Common Difference, d = - 9 + 12 = 3

t_{n} = a + (n - 1)d

⇒ 21 = - 12 + 3(n - 1)

⇒ 21 + 12 = 3(n - 1)

⇒ 3(n - 1) = 33

⇒ n - 1 = 11

∴n = 12

Number of terms of the series = 12

If 1 is added to each term of the above A.P.,

- 11, - 8, - 5,…….,22

Number of terms in the series, n_{1} = 12

Sum of all the terms,

The sum of the terms = 66

### Solution 38

Sum of n terms of the A.P., S_{n} = 3n^{2} + 6n

S_{1} = 3 × 1^{2} + 6 × 1 = 9 = t_{1} ……(i)

S_{2} = 3 × 2^{2} + 6 × 2 = 24 = t_{1} + t_{2} …….(ii)

S_{3} = 3 × 3^{2} + 6 × 3 = 45 = t_{1} + t_{2} + t_{3} ……..(iii)

From (i), (ii) and (iii),

t_{1 }= 9, t_{2} = 15, t_{3} = 21

Common difference, d = 15 - 9 = 6

n^{th} of the AP, t_{n} = a + (n - 1)d

= 9 + (n - 1) 6

= 9 + 6n - 6

= 6n + 3

Thus, the n^{th} term of the given A.P. = 6n + 3

### Solution 39

S_{n} = 5n - n^{2}

S_{1} = 5 × 1 - 1^{2} = 4 = t_{1}………..(i)

S_{2} = 5 × 2 - 2^{2} = 6 = t_{1} +t_{2}………..(ii)

S_{3} = 5 × 3 - 3^{2} = 6 = t_{1 }+ t_{2 }+ t_{3}……….(iii)

From (i), (ii) and (iii),

t_{1} = 4, t_{2} = 2, t_{3} = 0

Here a = 4, d = 2 - 4 = - 2

t_{n} = a + (n - 1)d

= 4 + (n - 1)( -2)

= 4 - 2n + 2

= 6 - 2n

### Solution 40

S_{n} = 4n^{2} + 2n

S_{1} = 4 × 1^{2} + 2 × 1 = 6 = t_{1}………….(i)

S_{2} = 4 × 2^{2} + 2 × 2 = 20 = t_{1} + t_{2}……….(ii)

S_{3} = 4 × 3^{2} + 2 × 3 = 42 = t_{1} + t_{2} + t_{3}………..(iii)

From (i), (ii) and (iii),

t_{1} = 6, t_{2} = 14, t_{3} = 22

Here a = 6, d = 14 - 6 = 8

t_{n} = a + (n - 1)d

t_{n} = 6 + (n - 1)8

= 6 + 8n - 8

= 8n - 2

*Note: Answer given in the book is incorrect.

### Solution 41

Sum of n terms of the A.P., S_{n} = 3n^{2} + 4n

S_{1} = 3 × 1^{2} + 4 × 1 = 7 = t_{1}………(i)

S_{2} = 3 × 2^{2} + 4 × 2 = 20 = t_{1} + t_{2}…….(ii)

S_{3} = 3 × 3^{2} + 4 × 3 = 39 = t_{1} + t_{2} + t_{3} …….(iii)

From (i), (ii), (iii)

t_{1} = 7, t_{2} = 13, t_{3} = 19

Common difference, d = 13 - 7 = 6

25^{th} of the term of this A.P., t_{25} = 7 + (25 - 1)6

= 7 + 144 = 151

∴The 25^{th} term of the A.P. is 151.

### Solution 42

Sum of the terms, S_{n} = 5n^{2} + 3n

S_{1} = 5 × 1^{2} + 3 × 1 = 8 = t_{1}………..(i)

S_{2} = 5 × 2^{2} + 3 × 2 = 26 = t_{1} + t_{2}…………..(ii)

S_{3} = 5 × 3^{2} + 3 × 3 = 54 = t_{1} + t_{2} + t_{3}…………(iii)

From(i), (ii) and (iii),

t_{1} = 8, t_{2} = 18, t_{3} = 28

Common difference, d = 18 - 8 = 10

t_{m} = 168

⇒ a + (m - 1)d = 168

⇒ 8 + (m - 1)×10 = 168

⇒ (m - 1) × 10 = 160

⇒ m - 1 = 16

∴m = 17

t_{20} = a + (20 - 1)d

= 8 + 19 × 10

= 8 + 190

= 198

### Solution 43

Let the first term be 'a' and the common difference be 'd'.

Sum of the first 'q' terms, S_{q} = 63q - 3q^{2}

S_{1} = 63 × 1 - 3 × 1^{2} = 60 = t_{1}……..(i)

S_{2} = 63 × 2 - 3 × 2^{2} = 114 = t_{1} + t_{2}…..(ii)

S_{3} = 63 × 3 - 3 × 3^{2} = 162 = t_{1} + t_{2 }+ t_{3} .....(iii)

From (i), (ii) and (iii),

t_{1} = 60

t_{2} = 54

t_{3} = 48

Common difference, d = 54 - 60 = - 6

t_{p} = a + (p - 1)d

⇒ -60 = 60 + (p - 1)( - 6)

⇒ - 120 = - 6(p - 1)

⇒ p - 1 = 20

∴p = 21

t_{11} = 60 + (11 - 1)( - 6)

=60 + 10( - 6)

= 60 - 60

= 0

The 11^{th} term of the A.P. is 0.

### Solution 44

Let the first term of the A.P. be 'a' and the common difference be 'd'

Sum of m terms of the A.P., S_{m} = 4m^{2} - m

S_{1} = 4 × 1^{2} - 1 = 3 = t_{1} …….(i)

S_{2} = 4 × 2^{2} - 2 = 14 = t_{1} + t_{2}……..(ii)

S_{3} = 4 × 3^{2} - 3 = 33 = t_{1 }+ t_{2} + t_{3} …….(iii)

From (i), (ii) and (iii)

t_{1} = 3, t_{2} = 11, t_{3} = 19

Common difference, d = 11 - 3 = 8

t_{n} = 107

⇒ a + (n - 1)d = 107

⇒ 3 + (n - 1)8 = 107

⇒ 8(n - 1) = 104

⇒ n - 1 = 13

∴n = 14

t_{21} = 3 + (21 - 1)8 = 3 + 160 = 163

### Solution 45

### Solution 46

### Solution 47 (i)

### Solution 47 (ii)

Sum of first n terms of an AP is given by

As per the question, S_{n} = n^{2}

Hence, the 10^{th} term of this A.P. is 19.

### Solution 48

### Solution 49

### Solution 50 (i)

### Solution 50 (ii)

### Solution 51

### Solution 52

### Solution 53

### Solution 54

### Solution 55 (i)

### Solution 55(ii)

### Solution 55(iii)

### Solution 55(iv)

### Solution 55(v)

### Solution 56 (i)

### Solution 56 (ii)

### Solution 56 (iii)

### Solution 56 (iv)

### Solution 56 (v)

### Solution 56 (vi)

### Solution 56(vii)

### Solution 56 (viii)

The nth term of an A.P. is given by a_{n} = a +
(n - 1)d

Sum of first n terms of an AP is given by

### Solution 57

### Solution 58

### Solution 59

### Solution 60

### Solution 61

Trees planted by the student in class 1 = 2 + 2 = 4

Trees planted by the student in class 2 = 4 + 4 = 8

Trees planted by the students in class 3 = 6 + 6 = 12

…….

Trees planted by the students in class 12 = 24 + 24 = 48

∴ the series will be 4, 8, 12,………., 48

a = 4, Common Difference, d = 8 - 4 = 4

Let 'n' be the number of terms in the series.

48 = 4 + (n - 1)4

⇒ 44 = 4(n - 1)

⇒ n - 1 = 11

∴n = 12

Sum of the A.P. series,

Number of trees planted by the students = 312

### Solution 62

Since, the difference between the savings of two consecutive months is Rs. 20, therefore the series is an A.P.

Here, the savings of the first month is Rs. 50

First term, a = 50, Common difference, d = 20

No. of terms = no. of months

No. of terms, n = 12

= 6[100 + 220]

= 6×320

= 1920

After a year, Ramakali will save Rs. 1920.

Yes, Ramakali will be able to fulfill her dream of sending her daughter to school.

### Solution 63

### Solution 64

### Solution 65

### Solution 66

### Solution 67

### Solution 68

### Solution 69

### Solution 70

Let the first term of the A.P. be 'a' and the common difference be 'd'.

R.H.S.

= 3(S_{20} - S_{10})

= 3(10[2a + 19d] - 5[2a + 9d])

= 3(20a + 190d - 10a - 45d)

= 3(10a + 145d)

= 3 × 5(2a + 29d)

= 15[2a + (30 - 1)d]

= S_{30}

= L.H.S.

### Solution 71

### Solution 72

### Solution 73

### Solution 74

## Arithmetic Progressions Exercise Ex. MCQs

### Solution 1

The nth term of an A.P. is given by a_{n} = a +
(n - 1)d

As per the question, a_{7} = 34 and a_{13}
= 64

Also,

Subtracting (i) from (ii), we get, 6d = 30

Therefore, d = 5

Substituting a in (i), we get, a = 4

The 18^{th} term is

a_{18} = a + 17d

= 4 + 85

= 89

Thus, the 18^{th} term is 89.

Hence, option (c) is correct.

### Solution 2

The sum of first n terms of an A.P. is given by

As per the question,

Subtracting (ii) from (i), we get

Now, sum of (p + q) terms is

Hence, option (d) is correct.

### Solution 3

Given: S_{n} = 3n^{2} + n and d = 6

Thus, the first term is 4.

Hence, option (d) is correct.

### Solution 4

Given: First term (a) = 1, last term (l) = 11 and S_{n}
= 36

We know that, sum of n terms is given by

Thus, the number of terms is 6.

Hence, option (b) is correct.

### Solution 5

Given: Sum of n terms S_{n} = 3n^{2} +
5n

For n = 1, we get

S_{1} = 3(1)^{2} + 5(1) = 8

For n = 2, we get

S_{2} = 3(2)^{2} + 5(2) = 22

We know that, a_{n} = S_{n} - S_{n -
1}

Therefore, we have

Common difference (d) = 14 - 8 = 6

Let 164 be the nth term of this A.P.

Thus, 164 is the 27^{th} term of this A.P.

Hence, option (b) is correct.

### Solution 6

Given: Sum of n terms S_{n} = 2n^{2} +
5n

For n = 1, we get

S_{1} = 2(1)^{2} + 5(1) = 7

For n = 2, we get

S_{2} = 2(2)^{2} + 5(2) = 18

We know that, a_{n} = S_{n} - S_{n -
1}

Here, common difference (d) = 11 - 7 = 4

The nth term is given by

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

= 7 + (n - 1)(4)

= 4n + 3

Thus, the nth term is 4n + 3.

Hence, option (c) is correct.

### Solution 7

Let (a - d), a and (a + d) be the first three consecutive terms of an A.P.

As per the question, we have

(a - d) + a + (a + d) = 51

i.e. 3a = 51

i.e. a = 17

Also, (a - d)(a + d) = 273

(a^{2} - d^{2}) = 273

289 - d^{2} = 273

d^{2} = 289 - 273

d^{2} = 16

d = ± 4

As the series is an increasing A.P., d must be positive.

Therefore, d = 4

So, we get, a + d = 21

Hence, option (c) is correct.

### Solution 8

Let (a - 3d), (a - d), (a + d) and (a + 3d) be the four numbers of an A.P.

As per the question, we have

(a - 3d) + (a - d) + (a + d) + (a + 3d) = 50

i.e. 4a = 50

i.e. a = 12.5

Also, a + 3d = 4(a - 3d)

i.e. a + 3d = 4a - 12d

i.e. 3a = 15d

i.e. a = 5d

i.e. 5d = 12.5

Therefore, d = 2.5

So, (a - 3d) = 5, (a - d) = 10, (a + d) = 15 and (a + 3d) = 20.

Thus, the numbers are 5, 10, 15 and 20.

Hence, option (a) is correct.

### Solution 9

Let a be the first term.

Sum of n terms of an A.P. is given by

So, the sum of (n - 1) terms is

And, sum of (n - 2) terms is

As it is given that d
= S_{n} - kS_{n-1} + S_{n-2}, we have

Hence, option (b) is correct.

### Solution 10

Sum of n terms of an A.P. is given by

Hence, option (b) is correct.

### Solution 11

The n even natural numbers 2, 4, 6,… forms an A.P. with first terms 2 and common difference 2.

So, the sum of first n even natural numbers is given by

The n odd natural numbers 1, 3, 5,… forms an A.P. with first terms 1 and common difference 2.

So, the sum of first n odd natural numbers is given by

As per the question,

S_{e} = k S_{o}

Hence, option (b) is correct.

### Solution 12

Given: First term = a, second term = b and last term = 2a

Therefore, common difference (d) = b - a

As the last term is 2a, we have

The of all the terms is given by

Hence, option (c) is correct.

### Solution 13

Sum of 'n' odd number of terms of an A.P. is

Therefore, we have

From equations (i) and (ii), we get

Hence, option (a) is correct.

### Solution 14

We know that, sum of n terms of an A.P. is given by

Also,

From (i) and (ii), we get

From (i), we get

2a = 2np - (n - 1)2p

i.e. 2a = 2np - 2np +2p

i.e. a = p

Now, the sum of p terms S_{p} is

Hence, option (c) is correct.

### Solution 15

We know that, sum of n terms of an A.P. is given by

As S_{2n} = 3S_{n}, we have

Consider,

Thus, S_{3n} : S_{n} = 6.

Hence, option (b) is correct.

### Solution 16

Given:

The sum of first n terms of an A.P. is given by

Subtracting (ii) from (i), we get

Therefore, S_{p+q} is

Hence, option (b) is correct.

### Solution 17

Given:

The sum of first n terms of an A.P. is given by

Therefore, S_{3n} will be

Similarly,

From (i), (ii) and (iii), we have

Thus,

Hence, option (c) is correct.

### Solution 18

Given: First term (a) = 2 and common difference (d) = 4

The sum of first n terms of an A.P. is given by

Sum of its 40 terms will be

Hence, option (a) is correct.

### Solution 19

In the given A.P. 3, 7, 11, 15, … we have

First term (a) = 3 and common difference (d) = 4

The sum of first n terms of an A.P. is given by

As the sum is given as 406, we have

Since the number of terms can't be negative, so n = 14.

Hence, option (d) is correct.

### Solution 20

For the given series, the terms forms an A.P.

Here, first term (a) = and common difference (d) =

The sum of first n terms of an A.P. is given by

Therefore, sum of the series is

Hence, option (c) is correct.

### Solution 21

In an A.P., a_{9} = 449 and a_{449} = 9

We know that the nth term is given by

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

Subtracting (ii) from (i), we get

-440d = 440

Therefore, d = -1

Putting the value of d in (i), we get

a - 8 = 449

Therefore, a = 457

Let nth term be 0 i.e. a_{n} = 0

Hence, option (c) is correct.

### Solution 22

As are in A.P.

Hence, option (c) is correct.

### Solution 23

Let the A.P. be a_{1}, a_{2}, a_{3},
…

As S_{n} is the sum of n terms of an A.P., we
have

S_{n} = a_{1} + a_{2} + a_{3}
+ … + a_{n}

∴ S_{n-1} = a_{1} + a_{2}
+ a_{3} + … + a_{n-1}

S_{n} - S_{n-1} = a_{1} + a_{2}
+ a_{3} + … + a_{n-1} + a_{n} - (a_{1} + a_{2}
+ a_{3} + … + a_{n-1})

= a_{n}

Thus, S_{n} - S_{n-1} = a_{n}.

Hence, option (b) is correct.

### Solution 24

Let the A.P. be a_{1}, a_{2}, a_{3},
…

When S_{n} is the sum of n terms of an A.P.,
nth term will be

a_{n} = S_{n} - S_{n-1}

Similarly,

a_{n-1} = S_{n-1} - S_{(n-1) - 1}
= S_{n-1} - S_{n-2}

Now, the common difference is given by

d = a_{n} - a_{n-1}

= S_{n}
- S_{n-1} - (S_{n-1} - S_{n-2})

= S_{n}
- 2S_{n-1} + S_{n-2}

Hence, option (a) is correct.

### Solution 25

Let S_{n} and S'_{n} denotes the sum of
first n terms of the two APs respectively.

Ratio of the sum of two APs is

Replacing n with (2n - 1), we get

Hence, option (b) is correct.

### Solution 26

As S_{n} is the sum of n terms of an A.P., we
have

As is independent of x, then

2a - d = 0

i.e. 2a = d

Putting this in the ratio, we get

Thus, 2a = d.

Hence, option (b) is correct.

### Solution 27

The n^{th} term of an A.P. is given by a_{n}
= a + (n - 1)d

Here, 'a' and 'd' are first term and common difference

As b is the n^{th} term of an A.P., we have

b = a + (n - 1)d

b - a = (n - 1)d

Hence, option (b) is correct.

### Solution 28

The n odd natural numbers are 1, 3, 5, … n.

These terms forms an A.P. with first term (a) = 1 and common difference (d) = 2.

So, the sum of first n odd natural numbers will be

Hence, option (c) is correct.

### Solution 29

Let d be the common difference of the two A.P.'s.

First term of 1^{st} A.P. (a) = 8

First term of 2^{nd} A.P. (a') = 3

Now, the 30^{th} term of 1^{st} A.P. is

a_{30} = a + (30 - 1)d

i.e. a_{30} = 8 + 29d … (i)

The 30^{th} term of 2^{nd} A.P. is

a'_{30} = a' + (30 - 1)d

a'_{30} = 3 + 29d … (ii)

Difference between these 30^{th} terms is

a_{30} - a'_{30} = (8 + 29d) - (3 +
29d) = 5

Hence, option (d) is correct.

### Solution 30

As 18, a, b, -3 are in A.P.

So, the difference between every two consecutive terms will be equal.

i.e. a - 18 = b - a = -3 - b

i.e. a - 18 = -3 - b

i.e. a + b = 18 - 3

i.e. a + b = 15

Hence, option (d) is correct.

### Solution 31

Let S_{n} and S'_{n} denotes the sum of
first n terms of the two APs respectively.

Ratio of the sum of two APs is

Replacing n with (2n - 1), we get

So, the ratio of their 18^{th} terms is

Hence, option (a) is correct.

### Solution 32

Here, 5, 9, 13, … forms an A.P. with first term (a) = 5 and common difference (d) = 4

Also, 7, 9, 11, … forms an A.P. with first term (a') = 7 and common difference (d') = 2

We know that, sum of n terms of an A.P. is given by

And,

From (i), we get

Since the number of terms can't be negative, so n = 7.

Hence, option (b) is correct.

### Solution 33

Let the A.P. be a_{1}, a_{2}, a_{3},
…

When S_{n} is the sum of n terms of an A.P.,
nth term will be

a_{n} = S_{n} - S_{n-1}

But, S_{n} = 3n^{2} + 5n

So, the n^{th} term will be

Let 164 be the nth term

Hence, option (b) is correct.

### Solution 34

The nth term of an A.P. is given as

a_{n} = 2n + 1

∴ First term = a_{1} = 2(1) +
1 = 3

a_{2} = 2(2) + 1 = 5

So, the common difference (d) = a_{2} - a_{1}
= 2

Now, the sum of n terms of an A.P. is given by

Hence, option (b) is correct.

### Solution 35

The nth term of an A.P. is given by

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

Ratio of 18^{th} and 11^{th} terms is

Now, the ratio of 21^{st} and 5^{th}
terms is

Hence, option (b) is correct.

### Solution 36

The n odd natural numbers are 1, 3, 5, … n.

These terms forms an A.P. with first term (a) = 1 and common difference (d) = 2.

So, the sum of first n odd natural numbers will be

Now, the sum of first 20 odd natural numbers is

Hence, option (c) is correct.

### Solution 37

The common difference of an A.P. with terms a_{1},
a_{2}, a_{3}, … is given by

d = a_{2} - a_{1} = a_{3} - a_{2}
= a_{4} - a_{3} = …

In the A.P. we have

Therefore,

Hence, option (a) is correct.

### Solution 38

The common difference of an A.P. with terms a_{1},
a_{2}, a_{3}, … is given by

d = a_{2} - a_{1} = a_{3} - a_{2}
= a_{4} - a_{3} = …

In the given A.P. we have

Therefore,

Hence, option (c) is correct.

### Solution 39

The common difference of an A.P. with terms a_{1},
a_{2}, a_{3}, … is given by

d = a_{2} - a_{1} = a_{3} - a_{2}
= a_{4} - a_{3} = …

In the given A.P. we have

Therefore,

Hence, option (d) is correct.

### Solution 40

Given: k, 2k - 1 and 2k + 1 are three consecutive terms of an AP

Therefore, common difference (d) = (2k - 1) - k

Also, d = (2k + 1) - (2k - 1)

∴ (2k - 1) - k = (2k + 1) - (2k - 1)

∴ k - 1 = 1 + 1

∴ k = 3

Hence, option (b) is correct.

### Solution 41

The given A.P. is

i.e.

Here, first term (a)

The common difference (d) =

As the A.P. is given by a, (a + d), (a + 2d), (a + 3d), …

So, the next term will be

(a + 3d) =

Hence, option (d) is correct.

### Solution 42

As the terms 3y - 1, 3y + 5 and 5y + 1 are in A.P.

Therefore, we have

3y + 5 - (3y - 1) = 5y + 1 - (3y + 5)

∴ 3y + 5 - 3y + 1 = 5y + 1 - 3y - 5

∴ 6 = 2y - 4

∴ y = 5

Hence, option (c) is correct.

### Solution 43

The first five multiples of 3 are 3, 6, 9, 12, 15.

These terms form an A.P. with,

First term (a) = 3 and common difference (d) = 3

We know that, sum of first n terms of an A.P. is

As sum of first 5 multiples is same as sum of first 5 terms of an A.P., we have

Sum of first five multiples = S_{5}

Hence, option (a) is correct.

### Solution 44

For the given A.P. 10, 6, 2, …, we have

First term (a) = 10 and common difference (d) = -4

We know that, sum of first n terms of an A.P. is given by

So, the sum of first 16 terms will be

Hence, option (a) is correct.

### Solution 45

In the A.P., we have

First term (a) = -5 and common difference (d) = 2

We know that, sum of first n terms of an A.P. is given by

So, the sum of first 6 terms will be

Hence, option (a) is correct.

### Solution 46

In the A.P., first term (a) = -11, last term (l) = 49 and common difference (d) = 3

Here, nth term from the end = l - (n - 1)d

Therefore,

4^{th} term from the end = 49 - (4 - 1)(3)

= 49 - 9

= 40

Hence, option (b) is correct.

### Solution 47

In the given A.P., first term (a) = 21 and common difference (d) = 21

Let 210 be the nth term of the AP.

So, we have

210 = a + (n - 1)d

210 = 21 + (n - 1)(21)

21(n - 1)= 189

n - 1 = 9

Therefore, n = 10

Hence, option (b) is correct.

### Solution 48

Given: 2^{nd} term = 13 and 5^{th} term
= 25

∴ a + (2 - 1)d = 13 and a + (5 - 1)d = 25

∴ a + d = 13 … (i) and,

a + 4d = 25 … (ii)

Subtracting (i) from (ii), we have

3d = 12

Therefore, d = 4

Substituting the value of 'd' in (i), we get

a = 9

So, the 7^{th} term will be

a_{7} = a + 6d = 9 + 24 = 33

Hence, option (b) is correct.

### Solution 49

As 2x, x + 10 and 3x + 2 are the three consecutive terms of an A.P.

We have

x + 10 - 2x = 3x + 2 - (x + 10)

∴ 10 - x = 2x - 8

∴ 3x = 18

Thus, x = 6

Hence, option (a) is correct.

### Solution 50

In the A.P., first term = p and common difference = q

The 10^{th} term will be

a_{10} = p + (10 - 1)q

= p + 9q

Hence, option (c) is correct.