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

Chapter 5 - 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)

Chapter 5 - 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)

Chapter 5 - 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

Chapter 5 - 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 16th term of the A.P.

To check whether -100 is any term of the A.P., take an 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,

12th 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, nth term is given by

an = a + (n - 1)d
an = 3 + (n - 1)4
an = 3 + 4n - 4
an = 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

 

tn = 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'

t24 = a + (24 - 1)d = a + 23d

t10­ = a + (10 - 1)d = a + 9d

t72 = a + (72 - 1)d = a + 71d

t15 = a + (15 - 1)d = a + 14d

 

t24 = 2t10

a + 23d = 2(a + 9d)

a + 23d = 2a + 18d

23d - 18d = 2a - a

5d = a

 

t72 = a + 71d

= 5d + 71d

= 76d

= 20d + 56d

= 4 × 5d + 4 × 14d

= 4(5d + 14d)

= 4(a + 14d)

= 4t15

t72 = 4t15

 

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 nth term of the series be 0.

tn = 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

an = 37

nth term of an A.P. is given by

a­n = 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 12th term.

a12 = 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

a5 = 31 and a25 = a5 + 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

Chapter 5 - 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 - d2 = 5d

d2 + 5d - 36 = 0

d2 + 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

Chapter 5 - 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

Sn = 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 S8 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.

   

tn = 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, S4 = 40 and S14 = 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.

tn = 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.

tn = 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.

tn = a + (n - 1)d

 

t10 = 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

 

tn = a + (n - 1)d

= 3 + (n - 1)2

= 3 + 2n - 2

tn= 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

 

28th term of the A.P., t28 = a + (28 - 1)d

= 3 + 27 × 2

= 3 + 54

= 57

 

The 28th 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'.

 

tn = - 4n + 15

t1 = - 4 × 1 + 15 = 11

t2 = - 4 × 2 + 15 = 7

t3 = - 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

 

tn = 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, n1 = 12

Sum of all the terms,

The sum of the terms = 66

Solution 38

Sum of n terms of the A.P., Sn = 3n2 + 6n

S1 = 3 × 12 + 6 × 1 = 9 = t1 ……(i)

S2 = 3 × 22 + 6 × 2 = 24 = t1 + t2 …….(ii)

S3 = 3 × 32 + 6 × 3 = 45 = t1 + t2 + t3 ……..(iii)

 

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

t1 = 9, t2 = 15, t3 = 21

 

Common difference, d = 15 - 9 = 6

nth of the AP, tn = a + (n - 1)d

= 9 + (n - 1) 6

= 9 + 6n - 6

= 6n + 3

 

Thus, the nth term of the given A.P. = 6n + 3

Solution 39

Sn = 5n - n2

S1 = 5 × 1 - 12 = 4 = t1………..(i)

S2 = 5 × 2 - 22 = 6 = t1 +t2………..(ii)

S3 = 5 × 3 - 32 = 6 = t1 + t2 + t……….(iii)

 

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

t1 = 4, t2 = 2, t3 = 0

 

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

tn = a + (n - 1)d

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

= 4 - 2n + 2

= 6 - 2n

Solution 40

Sn = 4n2 + 2n

S1 = 4 × 12 + 2 × 1 = 6 = t1………….(i)

S2 = 4 × 22 + 2 × 2 = 20 = t1 + t2……….(ii)

S3 = 4 × 32 + 2 × 3 = 42 = t1 + t2 + t3………..(iii)

 

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

t1 = 6, t2 = 14, t3 = 22

 

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

tn = a + (n - 1)d

tn = 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., Sn = 3n2 + 4n

S1 = 3 × 12 + 4 × 1 = 7 = t1………(i)

S2 = 3 × 22 + 4 × 2 = 20 = t1 + t…….(ii)

S3 = 3 × 32 + 4 × 3 = 39 = t1 + t2 + t3 …….(iii)

 

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

t1 = 7, t2 = 13, t­3 = 19

Common difference, d = 13 - 7 = 6

 

25th of the term of this A.P., t25 = 7 + (25 - 1)6

= 7 + 144 = 151

The 25th term of the A.P. is 151.

Solution 42

Sum of the terms, Sn = 5n2 + 3n

S1 = 5 × 12 + 3 × 1 = 8 = t1………..(i)

S2 = 5 × 22 + 3 × 2 = 26 = t1 + t2…………..(ii)

S3 = 5 × 32 + 3 × 3 = 54 = t1 + t2 + t3…………(iii)

 

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

t1 = 8, t2 = 18, t3 = 28

Common difference, d = 18 - 8 = 10

 

tm = 168

a + (m - 1)d = 168

8 + (m - 1)×10 = 168

(m - 1) × 10 = 160

m - 1 = 16

m = 17

t20 = 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, Sq = 63q - 3q2

S1 = 63 × 1 - 3 × 12 = 60 = t1……..(i)

S2 = 63 × 2 - 3 × 22 = 114 = t1 + t2…..(ii)

S3 = 63 × 3 - 3 × 32 = 162 = t1 + t2 + t3 .....(iii)

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

t1 = 60

t2 = 54

t3 = 48

 

Common difference, d = 54 - 60 = - 6

 

tp = a + (p - 1)d

-60 = 60 + (p - 1)( - 6)

- 120 = - 6(p - 1)

p - 1 = 20

p = 21

 

t11 = 60 + (11 - 1)( - 6)

=60 + 10( - 6)

= 60 - 60

= 0

The 11th 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., Sm = 4m2 - m

S1 = 4 × 12 - 1 = 3 = t1 …….(i)

S2 = 4 × 22 - 2 = 14 = t1 + t2……..(ii)

S3 = 4 × 32 - 3 = 33 = t1 + t2 + t …….(iii)

 

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

t1 = 3, t2 = 11, t3 = 19

Common difference, d = 11 - 3 = 8

 

tn = 107

a + (n - 1)d = 107

3 + (n - 1)8 = 107

8(n - 1) = 104

n - 1 = 13

n = 14

 

t21 = 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, Sn = n2

Hence, the 10th 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 an = 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(S20 - S10)

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

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

= 3(10a + 145d)

= 3 × 5(2a + 29d)

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

= S30

= L.H.S.

Solution 71

Solution 72

Solution 73

Solution 74

Chapter 5 - Arithmetic Progressions Exercise Ex. MCQs

Solution 1

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

As per the question, a7 = 34 and a13 = 64

Also,

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

Therefore, d = 5

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

The 18th term is

a18 = a + 17d

 = 4 + 85

 = 89

Thus, the 18th 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: Sn = 3n2 + 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 Sn = 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 Sn = 3n2 + 5n

For n = 1, we get

S1 = 3(1)2 + 5(1) = 8

For n = 2, we get

S2 = 3(2)2 + 5(2) = 22

We know that, an = Sn - Sn - 1

Therefore, we have

Common difference (d) = 14 - 8 = 6

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

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

Hence, option (b) is correct.

Solution 6

Given: Sum of n terms Sn = 2n2 + 5n

For n = 1, we get

S1 = 2(1)2 + 5(1) = 7

For n = 2, we get

S2 = 2(2)2 + 5(2) = 18

We know that, an = Sn - Sn - 1

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

The nth term is given by

an = 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

(a2 - d2) = 273

289 - d2 = 273

d2 = 289 - 273

d2 = 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 = Sn - kSn-1 + Sn-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,

Se = k So

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 Sp is

Hence, option (c) is correct.

Solution 15

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

As S2n = 3Sn, we have

Consider,

Thus, S3n : Sn = 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, Sp+q is

Hence, option (b) is correct.

Solution 17

Given:

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

Therefore, S3n 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., a9 = 449 and a449 = 9

We know that the nth term is given by

an = 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. an = 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 a1, a2, a3, …

As Sn is the sum of n terms of an A.P., we have

Sn = a1 + a2 + a3 + … + an

Sn-1 = a1 + a2 + a3 + … + an-1

Sn - Sn-1 = a1 + a2 + a3 + … + an-1 + an - (a1 + a2 + a3 + … + an-1)

 = an

Thus, Sn - Sn-1 = an.

Hence, option (b) is correct.

Solution 24

Let the A.P. be a1, a2, a3, …

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

an = Sn - Sn-1

Similarly,

an-1 = Sn-1 - S(n-1) - 1 = Sn-1 - Sn-2

Now, the common difference is given by

d = an - an-1

 = Sn - Sn-1 - (Sn-1 - Sn-2)

 = Sn - 2Sn-1 + Sn-2

Hence, option (a) is correct.

Solution 25

Let Sn 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 Sn 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 nth term of an A.P. is given by an = a + (n - 1)d

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

As b is the nth 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 1st A.P. (a) = 8

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

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

a30 = a + (30 - 1)d

i.e. a30 = 8 + 29d … (i)

The 30th term of 2nd A.P. is

a'30 = a' + (30 - 1)d

a'30 = 3 + 29d … (ii)

Difference between these 30th terms is

a30 - 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 Sn 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 18th 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 a1, a2, a3, …

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

an = Sn - Sn-1

But, Sn = 3n2 + 5n

So, the nth 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

an = 2n + 1

First term = a1 = 2(1) + 1 = 3

a2 = 2(2) + 1 = 5

So, the common difference (d) = a2 - a1 = 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

an = a + (n - 1)d

Ratio of 18th and 11th terms is

Now, the ratio of 21st and 5th 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 a1, a2, a3, … is given by

d = a2 - a1 = a3 - a2 = a4 - a3 = …

In the A.P.   we have

Therefore,   

Hence, option (a) is correct.

Solution 38

The common difference of an A.P. with terms a1, a2, a3, … is given by

d = a2 - a1 = a3 - a2 = a4 - a3 = …

In the given A.P.   we have

Therefore,   

Hence, option (c) is correct.

Solution 39

The common difference of an A.P. with terms a1, a2, a3, … is given by

d = a2 - a1 = a3 - a2 = a4 - a3 = …

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 = S5

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,

4th 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: 2nd term = 13 and 5th 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 7th term will be

a7 = 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 10th term will be

a10 = p + (10 - 1)q

 = p + 9q

Hence, option (c) is correct.