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

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.1

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.2

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.3

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.4

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.

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

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

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.

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}

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.

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

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.

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.

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, …..

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.5

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.

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.6

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.

Multiples of 8 are8,16,24,…

Now,

n=15, a=8, d=8

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

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

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

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

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.

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

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

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

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

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

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

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.

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,…….

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.

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

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

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

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.

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.

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

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.

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

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.

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

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

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.

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.

## Chapter 5 - Arithmetic Progressions Exercise Ex. MCQs

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.

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.

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

Thus, the first term is 4.

Hence, option (d) is correct.

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.

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.

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.

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.

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.

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.

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

Hence, option (b) is correct.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

As are in A.P.

Hence, option (c) is correct.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

### Other Chapters for CBSE Class 10 Mathematics

Chapter 1- Real Numbers Chapter 2- Polynomials Chapter 3- Pairs of Linear Equations in Two Variables Chapter 4- Quadratic Equations Chapter 6- Co-ordinate Geometry Chapter 7- Triangles Chapter 8- Circles Chapter 9- Constructions Chapter 10- Trigonometric Ratios Chapter 11- Trigonometric Identities Chapter 12- Heights and Distances Chapter 13- Areas Related to Circles Chapter 14- Surface Areas and Volumes Chapter 15- Statistics Chapter 16- Probability### RD SHARMA Solutions for CBSE Class 10 Subjects

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