# RD Sharma Solutions for Class 10 Maths Chapter 9 Arithmetic Progression MCQs and VSAQs

**VSAQs**

**1. Define an arithmetic progression.**

**Solution**

A sequence a_{1}, a_{2}, a_{3}, …, an is called an arithmetic progression of then exists a constant d

Such that a_{2} – a_{1} = d, a_{3} – a_{2} = d, …… a_{n} – a_{n-1} = d

and so on and d is called common difference

**2. Write the common difference of an A.P. whose nth term is a _{n} = 3n + 7.**

**Solution**

a_{n} = 3n + 7

a_{1} = 3×1 + 7 = 3 + 7 = 10

a_{2} = 3×2 + 7 = 6 + 7 = 13

a_{3} = 3×3 + 7 = 9 + 7 = 16

d = a_{3} – a_{2} or a_{2} – a_{1} = 16 – 13 = 3 or 13 – 10 = 3

**3. Which term of the sequence 114, 109, 104, … is the first negative term ?**

**Solution**

Sequence is 114, 109, 104, …..

Let a_{n} term be negative

**4. Write the value of a _{30} – a_{10} for the A.P. 4, 9, 14, 19, ……**

**Solution**

In the A.P. 4, 9, 14, 19,....

First term (a) = 4

Common difference (d) = 9 - 4 = 5

**5. Write 5th term from the end of the A.P. 3, 5, 7, 9,…, 201.**

**Solution**

A.P. is 3, 5, 7, 9, ..., 201

Here, first term (a) = 3

Common difference (d) = 5 - 3 = 2

= 3 + 190 = 193

5th term from the end = 193

**6. Write the value of x for which 2x, x + 10 and 3x + 2 are in A.P.**

**Solution**

2x, x+10 and 3x+2 are in A.P.

**7. Write the nth term of an A.P. the sum of whose n terms is S _{n}.**

**Solution**

Sum of n terms = S_{n}

Let a be the first term and d be the common difference a_{n} =S_{n} – S_{n-1}

**8. Write the sum of first n odd natural numbers.**

**Solution**

The first n odd natural numbers are 1, 3, 5, 7,....

Here, a = 1, d = 3 - 1 = 2

**9. Write the sum of first n even natural numbers.**

**Solution**

First n even natural numbers are

2, 4, 6, 8, ……

Here a = 2, d = 2

**10. If the sum of n terms of an A.P. is S _{n} = 3n² + 5n. Write its common difference.**

**Solution**

**11. Write the expression for the common difference of an A.P. Whose first term is a and nth term is b.**

**Solution**

First term of an A.P. = a

and a_{n} = a + (n – 1) d = b

Subtracting, b – a = (n – 1) d

**12. The first term of an A.P. is p and its common difference is q. Find its 10th term.**

**Solution**

First term of an A.P. (a) = p

and common difference (d) = q

a_{10} = a + (n – 1) d

= p + (10 – 1) q = p + 9q

**13. For what value of p are 2p + 1, 13, 5p – 3 are three consecutive terms of an A.P.?**

**Solution**

∵ 2p + 1, 13, 5p - 3 are consecutive terms of an A.P.

∴ c.d. = 13 - 2p - 1

⇒ 5p + 2p = 13 - 1 + 13 + 3

**14. If 4/5, a, 2 are three consecutive terms of an A.P., then find the value of a.**

**Solution**

**15. If the sum of first p term of an A.P. is ap² + bp, find its common difference.**

**Solution**

**16. Find the 9th term from the end of the A.P. 5, 9, 13, …, 185.**

**Solution**

Here, first term, a = 5

Common difference, d = 9 – 5 = 4

Last term, l = 185

nth term from the end = l – (n – 1) d

9th term from the end = 185 – (9 – 1) 4 = 185 – 8×4 = 185 – 32 = 153

**17. For what value of k will the consecutive terms 2k + 1, 3k + 3 and 5k – 1 form on A.P.?**

**Solution**

(3k + 3) – (2k + 1) = (5k – 1) – (3k + 3)

⇒ 3k + 3 – 2k – 1 = 5k – 1 – 3k – 3

⇒ k + 2 = 2k – 4

⇒ 2k – k = 2 + 4

⇒ k = 6

**18. Write the nth term of the A.P.1/m, 1+m/m, 1+2m/m,....**

**Solution**

### MCQs

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

(a) 87

(b) 88

(c) 89

(d) 90

**Solution**

**(c)**7th term (a

_{7}) = a + 6d = 34

13th term (a

_{13}) = a + 12d = 64

Subtracting, 6d = 30 => d = 5

and a + 12×5 = 64

⇒ a + 60 = 64

⇒ a = 64 – 60 = 4

18th term (a

_{18}) = a + 17d = 4 + 17×5 = 4 + 85 = 89

**2. 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)

**Solution**

**(d)**Sum of p terms = q

**3. 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

**Solution**

**(d)**Sum of n terms of an A.P. = 3n

^{2}+ n

and common difference (d) = 6

**4. 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

**Solution**

**(b)**First term of an A.P. (a) = 1

Last term (l) = 11

and sum of its terms = 36

Let n be the number of terms and d be the common difference, then

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

(a) 26th

(b) 27th

(c) 28th

(d) none of these

**Solution**

**(b)**Sum of n terms of an A.P. = 3n

^{2}+ 5n

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

**6. If the sum of it 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

**Solution**

**(c)**Let a be the first term and d be the common difference of an A.P. and

**7. 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

**Solution**

**(c)**

Let three consecutive terms of an increasing A.P. be a-d, a, a+d where a is the first term and d be the common difference.

a-d + a + a+d = 51

∴ d = ±4

∵ The A.P. is increasing.

∴ d = 4

Now, third term = a + d

= 17 + 4 = 21

**8. 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

**Solution**

**(a)**

4 numbers are in A.P.

Let the numbers be

a – 3d, a – d, a + d, a + 3d

Where a is the first term and 2d is the common difference

Now their sum = 50

a – 3d + a – d + a + d + a + 3d = 50

and the greatest number is 4 times the least number

a + 3d = 4 (a – 3d)

⇒ a + 3d = 4a – 12d

⇒ 4a – a = 3d + 12d

⇒ 3a = 15d

⇒ a = 5d

**9. Let S denotes 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} – k S_{n-1} + S_{n-2} then k =**

(a) 1

(b) 2

(c) 3

(d) None of these

**Solution**

**(b)**

S_{n} is the sum of n terms of an A.P.

a is its first term and d is common difference.

**10. 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:l ^{2} - a^{2}/k-(l+a) then k =**

(a) S

(b) 2S

(c) 3S

(d) None of these

**Solution**

**(b)**

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

(a) 1/n

(b) n-1/n

(c) n+1/2n

(d) n+1/n

**Solution**

**(d)**

Sum of n even natural number = n(n+1)

and sum of n odd natural numbers = n^{2}∴ n(n+1) = kn^{2}

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

(a) ab/2(b-a)

(b) ab/b-a

(c) 3ab/2(b-a)

(d) None of these

**Solution**

**(c)**First term (a

_{1}) = a

Second term (a

_{2}) = b

and last term (l) = 2a

∴ d = second term = first term = b-a

∵ l = an + (n-1)d

⇒ 2a = a + (n-1) (b-a)

⇒ (n-1)(b-a) = a

**13. If S _{1} is the sum of an arithmetic progression of ‘n’ odd number of terms and S_{2} is the sum of the terms of the series in odd places, then S_{1}/S_{2} =**

(a) 2n/n+1

(b) n/n+1

(c) n+1/2n

(d) n-1/n

**Solution**

**(a)**

Odd numbers are 1, 3, 5, 7, 9, 11, 13,.... n

S

_{1}= Sum of odd numbers = n

^{2}

S

_{2}= Sum of numbers at odd places 3, 7, 11, 15, ....

**14. If in an A.P., S**

_{n}= n^{2}p and S_{m}= m^{2}p, where S denotes the sum of r terms of the A.P., then S_{p}is equal to(a) 1/2 p^{3}

(b) mnp

(c) p^{3}

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

**Solution**

**(c)**

S_{n} = n^{2}p, S_{m} = m^{2}p

S_{r} = r^{2}p and S_{p} = p^{2}q = p^{3}

Hence, S_{p} = p^{3}

**15. 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

**Solution****(b)**

_{n}= Sum of n terms of an A.P. and S

_{2n}= 3S

_{n}

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

(a) 0

(b) – (p + q)

(c) p + q

(d) pq

**Solution**

**(c)** In an A.P. S_{p} = q, S_{q} = p

S_{p+q} = Sum of (p + q) terms = Sum of p term + Sum of q terms = q + p

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

(a) n

(b) 3n

(b) 3

(d) None of these

**Solution**

**(c)**