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

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.1

Write the first five terms for the sequence whose nth terms is:

a_{n} = n^{2} - n + 1

Find the indicated terms in the sequence whose nth terms are :

a_{n} = (n - 1) (2 - n) (3 + n); a_{1}, a_{2}, a_{3}

Find the indicated terms in the sequence whose nth terms are:

a_{n} = (-1)^{n} n; a_{3}, a_{5}, a_{8}

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.2

Justify whether it is true to say that the sequence having following nth term is an A.P.

a_{n} = 2n - 1

Justify whether it is true to say that the sequence having following nth term is an A.P.

a_{n} = 3n^{2} + 5

Justify whether it is true to say that the sequence having following nth term is an A.P.

a_{n} = 1 + n + n^{2}

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.3

Write the arithmetic progression when first term a and common difference d are as follows:

(i) a = 4, d = -3

(ii) a = -1, d = 1/2

(iii) a = -1.5, d = -0.5

In which of the following situations, the sequence of numbers formed will form an A.P.?

The cost of digging a well for the first metre is Rs 150 and rises by Rs 20 for each succeeding metre.

In which of the following situations, the sequence of numbers formed will form an A.P.?

The amount of air present in the cylinder when a vacuum pump removes each time 1/4 of their remaining in the cylinder.

In which of the following situations, the sequence of numbers formed will form an A.P.?

Divya deposited Rs.1000 at compound interest at the rate of 10% per annum. The amount at the end of first year, second year, third year, …., and so on.

Find out whether of the given sequence is an arithemtic progressions. If it is an arithmetic progressions, find out the common difference.

p, p + 90, p + 180, p + 270, ... where p = (999)^{999}

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.4

Which term of the A.P. 4, 9, 14, ... is 254?

Which term of the A.P. -7, -12, -17, -22,… will be -82? Is -100 any term of the A.P.?

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.

Find the 12^{th} term from the end of the following arithmetic progressions:

3, 8, 13, ..., 253

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

The 26^{th}, 11^{th} and last term of an A.P. are 0, 3 and , respectively. Find the common difference and the number of terms.

The sum of 4^{th} and 8^{th} terms of an A.P. is 24 and the sum of the 6^{th} and 10^{th} terms is 34. Find the first term and the common difference of the A.P.

The eighth term of an A.P. is half of its second term and the eleventh term exceeds one third of its fourth term by 1. Find the 15^{th} term.

The 17^{th} term of an A.P. is 5 more than twice its 8^{th} term. If the 11^{th} term of the A.P. is 43, find the n^{th} term.

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

Find the number of all three digit natural numbers which are divisible by 9.

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.

The 19^{th} term of an A.P. is equal to three times its sixth term. If its 9^{th} term is 19, find the A.P.

The 9^{th} term of an A.P. is equal to 6 times its second term. If its 5^{th} term is 22, find the A.P.

The 24^{th} term of an A.P. is twice its 10^{th} term. Show that its 72^{nd} term is 4 times its 15^{th} term.

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}

Find the number of natural numbers between 101 and 999 which are divisible by both 2 and 5.

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.

If the seventh term of an A.P. is 1/9 and its ninth term is 1/7, find its (63)^{rd} term.

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

The sum of 5^{th} and 9^{th} terms of an A.P. is 30. If its 25^{th} term is three times its 8^{th} term, find the A.P.

Find where 0(zero) is a term of the A.P. 40, 37, 34, 31, ...

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.

Find the middle term of the A.P. 213, 205, 197, …, 37.

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.

If the 5^{th} term of an A.P. is 31 and 25^{th} term is 140 more than the 5^{th} term, find the A.P.

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, …..

Find the sum of two middle terms of the A.P.

How many numbers lie between 10 and 300, which when divided by 4 leave a remainder 3?

Find the 12^{th} term from the end of the A.P. -2, -4, -6, …, -100.

For the A.P.: -3, -7, -11, …, can we find a_{30} - a_{20} without actually finding a_{30} and a_{20}? Give reason for your answer.

Two A.P.s have the same common difference. The first term of one A.P. is 2 and that of the other is 7. The difference between their 10^{th} terms is the same as the difference between their 21^{st} terms, which is the same as the difference between any two corresponding terms. Why?

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.5

Divide 56 in four parts in A.P. such that the ratio of the product of their extremes to the product of their means is 5 : 6.

The sum of the first three numbers in an arithmetic progression is 18. If the product of the first and third term is 5 times the common difference, find the three numbers.

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.

Spilt 207 into three parts such that these are in A.P. and the product of the two smaller parts is 4623.

The angles of a triangle are in A.P. the greatest angle is twice the least. Find all the angles.

The sum of four consecutive numbers in A.P. is 32 and the ratio of the product of the first and last terms to the product of two middle terms is 7 : 15. Find the number.

## Chapter 5 - Arithmetic Progressions Exercise Ex. 5.6

Find the sum of last ten terms of the A.P: 8, 10, 12, 14,.., 126.

Find the sum of the first 15 terms of each of the following sequences having *n*^{th} term as* y _{n}* = 9 - 5

*n*

How many terms are there in the A.P. whose first and fifth terms are -14 and 2 respectively and the sum of the terms is 40?

How many terms of the A.P. 9, 17, 25, ... must be taken so that their sum is 636?

Remark* - Question modified.

How many terms of the A.P. 27, 24, 21… should be taken so that their sum is zero?

How many terms of the A.P. 45, 39, 33 … must be taken so that their sum is 180? Explain the double answer.

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.

Find the sum of

The first 15 multiples of 8.

Multiples of 8 are8,16,24,…

Now,

n=15, a=8, d=8

Find the sum of

The first 40 positive integers divisible by (a) 3 (b) 5 (c) 6.

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

Find the sum of

All 3 - digit natural numbers which are divisible by 13.

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

Find the sum of

All 3 - digit natural numbers, which are multiples of 11.

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

Find the sum of

All 2 - digit natural numbers divisible by 4.

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

Find the sum of first 8 multiples of 3.

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.

Find the sum:

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

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

If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

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

The first and the last terms of an A.P. are 7 and 49 respectively. If sum of all its terms is 420, find its common difference.

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

The first and the last terms of an A.P. are 5 and 45 respectively. If the sum of all its terms is 400, find its common difference.

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

The sum of first q terms of an A.P. is 162. The ratio of its 6^{th} term to its 13^{th} term is 1 : 2. Find the first and 15^{th} term of the A.P.

If the 10^{th} term of an A.P. is 21 and the sum of its first ten terms is 120, find its n^{th} term.

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

The sum of the first 7 terms of an A.P. is 63 and the sum of its next 7 terms is 161. Find the 28^{th} term of this A.P.

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.

The sum of first seven terms of an A.P. is 182. If its 4^{th} and the 17^{th} terms are in the ratio 1: 5, find the A.P.

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,…….

The n^{th} term of an A.P. is given by (- 4n + 15). Find the sum of the first 20 terms of this A.P.

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.

Sum of the first 14 terms of an A.P. is 1505 and its first term is 10. Find its 25^{th} term.

Find the number of terms of the A.P. - 12, - 9, - 6, …, 21. If 1 is added to each term of this A.P., then find the sum of all terms of the A.P. thus obtained.

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

The sum of the first n terms of an A.P. is 3n^{2} + 6n. Find the nth term of this A.P.

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

The sum of the first n terms of an A.P. is 5n - n^{2}. Find the n^{th} term of this A.P.

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

The sum of the first n terms of an A.P. is 4n^{2} + 2n. find the n^{th} term of this A.P.

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.

The sum of first n terms of an A.P. is 3n^{2} + 4n. find the 25^{th} term of this A.P.

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.

The sum of first n terms of an A.P. is 5n^{2} + 3n. If its m^{th} term is 168, find the value of m. Also, find the 20^{th} term of this A.P.

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

The sum of first q terms of an A.P. is 63q - 3q^{2}. If its pth term -60, find the value of p. Also, find the 11^{th} term of this A.P.

Remark* - Question modified.

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.

The sum of first m terms of an A.P. is 4m^{2} - m. If its nth term is 107, find the value of n. Also, find the 21^{st} term of this A.P.

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

If the sum of first n terms of an A.P. is then find its n^{th} term. Hence write its 20^{th} term.

If
the sum of first n terms of an A.P. is n^{2}, then find its 10^{th}
term.

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.

Find the sum of all integers between 100 and 550 which are not divisible by 9.

Find the sum of all integers between 1 and 500 which are multiplies 2 as well as of 5.

Find the sum of all integers from 1 to 500 which are multiplies 2 as well as of 5.

Find the sum of all integers from 1 to 500 which are multiples of 2 or 5.

Let there be an A.P. with first term 'a', common difference 'd'. If a_{n} denotes its n^{th} term and S_{n} the sum of first n terms, find

n and a_{n}, if a = 2, d = 8 and S_{n} = 90.

Let there be an A.P. with first term 'a', common difference 'd'. If a_{n} denotes its n^{th} term and S_{n} the sum of first n terms, find k, if S_{n} = 3n^{2} + 5n and a_{k} = 164.

Let
there be an A.P. with first term 'a', common difference 'd'. If a_{n}
denotes its n^{th} term and S_{n} the sum of first n terms,
find S_{22}, if d = 22 and a_{22} = 149.

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

If S_{n} denotes the sum of first n terms of an A.P., prove that S_{12} = 3(S_{8} - S_{4}).

A thief, after committing a theft runs at a uniform speed of 50 m/minute. After 2 minutes, a policeman runs to catch him. He goes 60 m in first minute and increases his speed by 5 m/minute every succeeding minute. After how many minutes, the policeman will catch the thief?

The sums of first n terms of three A.P.s are S_{1}, S_{2} and S_{3}. The first term of each is 5 and their common differences are 2, 4 and 6 respectively. Prove that S_{1} + S_{3} = 2S_{2}.

Resham wanted to save at least Rs. 6500 for sending her daughter to school next year (after 12 months). She saved Rs. 450 in the first month and raised her saying by Rs.20 every next month. How much will she be able to save in next 12 months? Will she be able to send her daughter to the school next year?

In a school, students decided to plant trees in and around the school to reduce air pollution. It was decided that the number of trees, that each section of each class will plant, will be double of the class in which they are studying. If there are 1 to 12 classes in the school and each class has two sections, find how many trees were planted by the students.

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

Ramkali would need Rs. 1800 for admission fee and books etc., for her daughter to start going to school from the next year. She saved Rs. 50 in the first month of this year and increased her monthly saving by Rs. 20. After a year, how much money will she save? Will she be able to fulfill her dream of sending her daughter to school?

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.

If S_{n} denotes the sum of the first n terms of an A.P., prove that S_{30} = 3(S_{20} - S_{10}).

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.

Solve the equation

(-4) + (-1) + 2 + 5 + …. + x = 437.

Which term of the A.P. -2, -7, -12,…, will be -77? Find the sum of this A.P. upto the term -77.

The sum of the first n terms of an A.P. whose first term is 8 and the common difference is 20 is equal to the sum of first 2n terms of another A.P. whose first term is -30 and common difference is 8. Find n.

The students of a school decided to beautify the school on the annual day by fixing colourful flags on the straight passage of the school. They have 27 flags to be fixed at intervals of every 2 metre. The flags are stored at the position of the middle most flag. Ruchi was given the responsibility of placing the flags. Ruchi kept her books where the flags were stored. She could carry only one flag at a time. How much distance did she cover in completing this job and returning back to collect her books? What is the maximum distance she travelled carrying a flag?

## Chapter 5 - Arithmetic Progressions Exercise Ex. MCQs

If
7^{th} and 13^{th} terms of an A.P. be 34 and 64
respectively, then its 18^{th} term is

(a) 87

(b) 88

(c) 89

(d) 90

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.

If the sum of p terms of an A.P. is q and the sum of q terms is p, then the sum of p + q terms will be

(a) 0

(b) p - q

(c) p + q

(d) -(p + q)

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.

If
the sum of n terms of an A.P. be 3n^{2} + n and its common difference
is 6, then its first term is

(a) 2

(b) 3

(c) 1

(d) 4

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

Thus, the first term is 4.

Hence, option (d) is correct.

The first and last terms of an A.P. are 1 and 11. If the sum of its terms is 36, then the number of terms will be

(a) 5

(b) 6

(c) 7

(d) 8

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.

If
the sum of n terms of an A.P. is 3n^{2} + 5n then which of its terms
is 164?

(a) 5

(b) 6

(c) 7

(d) 8

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.

If
the sum of n terms of an A.P. is 2n^{2} + 5n, then its nth term is

(a) 4n - 3

(b) 3n - 4

(c) 4n + 3

(d) 3n + 4

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.

If the sum of three consecutive terms of an increasing A.P. is 51 and the product of the first and third of these terms is 273, then the third term is

(a) 13

(b) 9

(c) 21

(d) 17

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.

If four numbers in A.P. are such that their sum is 50 and the greatest number is 4 times the least, then the numbers are

(a) 5, 10, 15, 20

(b) 4, 10, 16, 22

(c) 3, 7, 11, 15

(d) none of these

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
S_{n} denote the sum of n terms of an A.P. whose first term is a. If
the common difference d is given by d = S_{n} - kS_{n-1} + S_{n-2},
then k =

(a) 1

(b) 2

(c) 3

(d) none of these

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.

The first and last term of an A.P. are a and l respectively. If S is the sum of all the terms of the A.P. and the common difference is given by then k =

(a) S

(b) 2S

(c) 3S

(d) none of these

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

Hence, option (b) is correct.

If the sum of first n even natural numbers is equal to k times the sum of first n odd natural numbers, then k =

(a)

(b)

(c)

(d)

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.

If the first, second and last terms of an A.P. are a, b and 2a respectively, its sum is

(a)

(b)

(c)

(d) none of these

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.

If
S_{1} is the sum of an arithmetic progression of 'n' odd number of
terms and S_{2} the sum of the terms of the series in odd places,
then

(a)

(b)

(c)

(d)

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.

If
in an A.P., S_{n} = n^{2}p and S_{m} = m^{2}p,
where S_{r} denotes the sum of r terms of the A.P., then S_{p}
is equal to

(a)

(b) mnp

(c) p^{3}

(d) (m + n)p^{2}

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.

If
S_{n} denote the sum of the first n terms of an A.P. If S_{2n}
= 3S_{n}, then S_{3n} : S_{n} is equal to

(a) 4

(b) 6

(c) 8

(d) 10

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.

In
an AP, S_{p} = q, S_{q} = p and S_{r} denotes the sum
of first r terms. Then, S_{p+q} is equal to

(a) 0

(b) -(p + q)

(c) p + q

(d) pq

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.

If
S_{r} denotes the sum of the first r terms of an A.P. Then, S_{3n}
: (S_{2n} - S_{n}) is

(a) n

(b) 3n

(c) 3

(d) none of these

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.

If the first term of an A.P. is 2 and common difference is 4, then the sum of its 40 terms is

(a) 3200

(b) 1600

(c) 200

(d) 2800

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.

The number of terms of the A.P. 3, 7, 11, 15, … to be taken so that the sum is 406 is

(a) 5

(b) 10

(c) 12

(d) 14

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.

Sum of n terms of the series is

(a)

(b) 2n(n + 1)

(c)

(d) 1

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.

The
9^{th} term of an A.P. is 449 and 449^{th} term is 9. The
term which is equal to zero is

(a) 501^{th}

(b) 502^{th}

(c) 458^{th}

(d) None of these

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.

If are in A.P. Then, x =

(a) 5

(b) 3

(c) 1

(d) 2

As are in A.P.

Hence, option (c) is correct.

The
nth term of an A.P., the sum of whose n terms is S_{n}, is

(a) S_{n} + S_{n-1}

(b) S_{n}
- S_{n-1}

(c) S_{n} + S_{n+1}

(d) S_{n}
- S_{n+1}

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.

The
common difference of an A.P., the sum of whose n terms is S_{n}, is

(a) S_{n} - 2S_{n-1} + S_{n-2}

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

(c) S_{n} - S_{n-2}

(d) S_{n}
- S_{n-1}

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.

If
the sums of n terms of two arithmetic progressions are in the ratio then their n^{th}
terms are in the ratio

(a)

(b)

(c)

(d)

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.

If
S_{n} denote the sum of n terms of an A.P. with first term a and
common d such that is independent of x,
then

(a) d = a

(b) d = 2a

(c) a = 2d

(d) d = -a

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.

If
the first term of an A.P. is a and n^{th} term is b, then its common
difference is

(a)

(b)

(c)

(d)

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 sum of first n odd natural numbers is

(a) 2n - 1

(b) 2n + 1

(c) n^{2}

(d) n^{2} - 1

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.

Two A.P.'s have the same common
difference. The first term of one of these is 8 and that of the other is 3.
The difference between their 30^{th} terms is

(a) 11

(b) 3

(c) 8

(d) 5

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.

If 18, a, b, -3 are in A.P., then a + b =

(a) 19

(b) 7

(c) 11

(d) 15

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.

The sum of n terms of two A.P.'s are
in the ratio 5n+9:9n+6. Then, the ratio of their 18^{th} term is

(a)

(b)

(c)

(d)

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.

If then n =

(a) 8

(b) 7

(c) 10

(d) 11

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.

The sum of n terms of an A.P. is 3n^{2}
+ 5n, then 164 is its

(a) 24^{th}
term

(b) 27^{th} term

(c) 26^{th} term

(d) 25^{th} term

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.

If the nth term of an A.P. is 2n + 1, then the sum of first n terms of the A.P. is

(a) n(n - 2)

(b) n(n + 2)

(c) n(n + 1)

(d) n(n - 1)

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.

If 18^{th} and 11^{th}
term of an A.P. are in the ratio 3 : 2, then its 21^{st} and 5^{th}
terms are in the ratio

(a) 3 : 2

(b) 3 : 1

(c) 1 : 3

(d) 2 : 3

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 sum of first 20 odd natural numbers is

(a) 100

(b) 210

(c) 400

(d) 420

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 the A.P. is

(a) -1

(b) 1

(c) q

(d) 2q

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 the A.P. is

(a)

(b)

(c) -b

(d) b

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 the A.P. is

(a) 2b

(b) -2b

(c) 3

(d) -3

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.

If k, 2k - 1 and 2k + 1 are three consecutive terms of an AP, the value of k is

(a) -2

(b) 3

(c) -3

(d) 6

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 next term of the A.P.

(a)

(b)

(c)

(d)

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.

The first three terms of an A.P. respectively are 3y - 1, 3y + 5 and 5y + 1. Then, y equals

(a) -3

(b) 4

(c) 5

(d) 2

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 sum of first five multiples of 3 is

(a) 45

(b) 55

(c) 65

(d) 75

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.

The sum of first 16 terms of the A.P.: 10, 6, 2 …, is

(a) -320

(b) 320

(c) -352

(d) -400

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.

If the first term of an A.P. is -5 and the common difference is 2, then the sum of first 6 terms is

(a) 0

(b) 5

(c) 6

(d) 15

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.

The 4^{th} term from the end
of the AP: -11, -8, -5, …, 49 is

(a) 37

(b) 40

(c) 43

(d) 58

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.

Which term of the A.P. 21, 42, 63, 84, … is 210?

(a) 9^{th}

(b) 10^{th}

(c) 11^{th}

(d) 12^{th}

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.

If the 2^{nd} term of an A.P.
is 13 and 5^{th} term is 25, what is its 7^{th} term?

(a) 30

(b) 33

(c) 37

(d) 38

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.

The value of x for which 2x, x + 10 and 3x + 2 are the three consecutive terms of an A.P; is

(a) 6

(b) -6

(c) 18

(d) -18

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.

The first term of an A.P. is p and the
common difference is q, then its 10^{th} term is

(a) q + 9p

(b) p - 9q

(c) p + 9q

(d) 2p + 9q

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

### Kindly Sign up for a personalised experience

- Ask Study Doubts
- Sample Papers
- Past Year Papers
- Textbook Solutions

#### Sign Up

#### Verify mobile number

Enter the OTP sent to your number

Change