Please refer to Assignments Class 10 Mathematics Arithmetic Progression Chapter 5 with solved questions and answers. We have provided Class 10 Mathematics Assignments for all chapters on our website. These problems and solutions for Chapter 5 Arithmetic Progression Class 10 Mathematics have been prepared as per the latest syllabus and books issued for the current academic year. Learn these solved important questions to get more marks in your class tests and examinations.

## Arithmetic Progression Assignments Class 10 Mathematics

**VERY SHORT ANSWER TYPE QUESTIONS**

**Question. Find 5th term of an A.P. whose n ^{th} term is 3n – 5**

**Ans.**a

_{n}= 3

_{n}– 5, a

_{5}= 10

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

**Ans.**a

_{1}= 3 + 7 = 10, a

_{2}= 6 + 7 = 13, d = 3

**Question. Find the sum of first 10 even numbers.****Ans.** S_{n} = 10/2 [2 × 2 + 9 × 2] = 110

**Question. 3, k – 2, 5 are in A.P. find k.****Ans.** 3, k – 2, 5 are in A.P.

∴ k – 2 = (3+5)/2 = 4 ,k = 6

**Question. What will be the value of a _{8} – a_{4} for the following A.P. 4, 9, 14, …………., 254**

**Ans.**(a + 7d) – (a + 3d) = 4d = 20

**Question. Write the sum of first n natural numbers.****Ans.** 1 + 2 + …….. + n = n/2 [1 + n]

**Question. 11th term of an A.P. – 3, -(1/2) , , … is**

(A) 28

(B) 22

(C) –38

(D) -48(1/2)**Ans. **(B) 22

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

(A) a + 9p

(B) p – 9q

(C) p + 9q

(D) 2p + 9q

**Ans.**(C) p + 9q

**Question. The famous mathematician associated with finding the sum of the first 100 natural numbers is**

(A) Pythagoras

(B) Newton

(C) Gauss

(D) Euclid**Ans. **(C) Gauss

**Question. The first three terms of an A.P. respectively are 3y – 1, 3y + 5 and 5y + 1**

(A) –3

(B) 4

(C) 5

(D) 2**Ans.** (C) 5

**Question. The nth term of the A.P. (1+√3) , (1+2√3), (1+3√3), …. is**

(A) 1 + n√3

(B) n + √3

(C) n(1+√3)

(D) n√3**Ans. **(A) 1 + n√3

**Question. Match the following :**

**Ans.** (a) → (b), (b) → (a), (c) → (e)

(d) → (c), (e) → (d)

**Question. n ^{th} term of an A.P. 5, 10, 15, 20 …. n terms and n^{th} term of A.P. 15, 30, 45, 60, … n terms are same. True or False.**

**Ans.**False, (If a, b, c, d … are in AP then ka, kb, kc, kd ….. are in AP)

k ≠ 0, n

^{th}term = k times n

^{th}term of original A.P. of new A.P.

**Question. Difference of m ^{th} and n^{th} term of an A.P. = (m – n) d. True or False.**

**Ans.**True, [a + (m – 1) d] – [a + (n – 1) d] = (m – n) d

**Question. Sum of first 20 natural numbers is 410. True or False.****Ans.** False, ∵ S_{n} = (n(n+1))/2 = (20×21)/2 = 210

**SHORT ANSWER TYPE QUESTIONS-I**

**Question. Which term of the A.P. 5, 15, 25, ……. will be 130 more than its 31 ^{st} term?**

**Ans.**Let a

_{n}= 130 + a

_{31}

Solve to get n = 44

∴ Answer is 44th term

**Question**. Which term of A.P. 3, 7, 11, 15 …. is 79? Also find the sum 3 + 7 + 11 + … + 79.**Ans.**

**Question. If 10 times of 10 ^{th} term is equal to 20 times of 20^{th} term of an A.P. Find its 30th term.**

**Ans.**ATQ 10 a

_{10}= 20 a

_{20}

⇒ a

_{10}= 2a

_{20}

a + 9d = 2a + 38d

a = – 29d …(1)

a30 = a + 29d

Substitute a from (1)

∴ Answer is a

_{30}= 0

**Question. Find the 20th term from the last term of the A.P. 3, 8, 13, … 253.****Ans.** 20th term from end using [l – (n – 1) d]

= 253 – 19 × 5

= 253 – 95 = 158

**Question**. Solve 1 + 4 + 7 + 10 + … + x = 287**Ans.**

**Question. Find how many two digit numbers are divisible by 6?****Ans.** Two digit numbers divisible by 6 are 12, 18, 24, …. 96.

a_{2} – a_{1} = a_{3} – a_{2} = 6

∴ A.P., a_{n} = 96 ⇒ n = 15

**Question. Which term of the A.P. : 121, 117, 113 … is the first negative terms ?****Ans.** Let a_{n} < 0

121 + (n – 1) (– 4) < 0

121 – 4n + 4 < 0

125 < 4n

n > 125/4

∴ n = 32

32nd term will be first negative term.

**Question. If 1/(x + 2), 1/(x + 3) and 1/(x + 5) are in A.P. find x.****Ans.** 2/(x+3) = 1/(x+2) + 1/(x+5) (2b = a + c)

Solve to get x = 1.

**Question**. In an A.P. find S_{n}, where an = 5_{n} – 1. Hence find the sum of the first 20 terms.**Ans. **

**SHORT ANSWER TYPE QUESTIONS-II**

**Question. In a flower bed, there are 23 rose plants in the first row, 21 in the second, 19 in the third and so on. There are 5 rose plants in the last row. How many rows are there in the flower bed?****Ans.** 23, 21, 19, … 5

a_{n} = a + (n – 1) d

5 = 23 + (n – 1) (– 2)

n = 10

**Question**. Find the sum of integers between 10 and 500 which are divisible by 7.**Ans.** Numbers between 10 and 500 which are divisible by 7, 14, 21, 28 …, 497

Find n, using a_{n} = a + (n – 1) d , then use S_{n} = n/2 [2a + (n – 1) d]

∴ Answer is S_{n} = 17885. (n = 70)

**Question. For what value of n, are the n ^{th} term of two A.P’s 63, 65, 67 ……… and 3, 10, 17 ….. are equal ?**

**Ans.**63, 65, 67, …..

a

_{n}= 63 + (n – 1) 2 = 61 + 2n

3, 10, 17, ….

a

_{n}= 3 + (n – 1) 7

= 7n – 4

61 + 2n = 7n – 4

65 = 5n

n = 13

**Question. If the pth term A.P. is q and the q ^{th} term is p, prove that its n^{th} term is (p + q – n).**

**Ans.**a

_{p}= q, a

_{q}= p

Solve to get a and d, a = q + p – 1, d = – 1

a

_{n}= p + q – n

**Question. 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, then find the n^{th} term of the A.P.**

**Ans.**ATQ,

a

_{17}= 5 + 2 × a

_{8}

a + 16d = 5 + 2a + 14 d

a – 2d = – 5 …(1)

a

_{11}= a + 10d = 43 ….(2)

Solving (1) & (2), we get

a = 3, d = 4

∴ a

_{n}= 4n – 1

**Question. 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.**

**Ans.**ATQ a

_{5}+ a

_{9}= 30

a

_{25}= 3 a

_{8}

Solve to get a = 3, d = 2

A.P. 3, 5, 7, 9, …

**Question. If S _{n}, the sum of first n terms of an A.P. is given by S_{n} = 3n^{2} – 4n, find the n^{th} term.**

**Ans.**S

_{n}= 3n

^{2}– 4n

a

_{n}= S

_{n}– S

_{n–1}

= (3n

^{2}– 4n) – [3(n – 1)

^{2}– 4(n – 1)]

= (3n

^{2}– 4n) – [3n

^{2}+ 3 – 6n – 4n + 4]

= – [7 – 6n]

a

_{n}= 6n – 7

**Question**. If the sum of the first 14 terms of an A.P. is 1050 and its fourth term is 40, find its 20^{th} term.**Ans.**

**Question**. If the m^{th} term of an A.P. be 1/n and n^{th} term be 1/m, show that its (mn)^{th} is 1.**Ans.**

**LONG ANSWER TYPE QUESTIONS**

**Question. The sum of first 20 terms of an A.P. is one third of the sum of next 20 term. If first term is 1, find the sum of first 30 terms of this A.P.****Ans.** ATQ S_{20} = 1/3 (S_{40} – S_{20}), a = 1

Use S_{n} = n/2 [2a + (n – 1) d] and a = 1 to find d, d = 2

then find S_{30}

∴ Answer is 900

**Question**. If the sum of the first four terms of an AP is 40 and the sum of the first fourteen terms of an AP is 280. Find the sum of first n terms of the A.P.**Ans.**

**Question**. The sum of third and seventh terms of an A.P. is 6 and their product is 8. Find the sum of first 16 terms of the A.P.**Ans.**

**Question**. In an AP of 50 terms, the sum of first 10 terms is 210 and the sum of last 15 terms is 2565. Find the A.P.**Ans.**

**Question. Determine the A.P. whose 4 ^{th} term is 18 and the difference of 9^{th} term from the 15^{th} term is 30.**

**Ans.**ATQ a

_{4}= 18 …(1), a

_{15}– a

_{9}= 30 …(2)

equation (2) will give d = 5

Substitute d = 5 in (1) to get a = 3

A.P. 3, 8, 13, ….

**Question. If the sum of the first seven terms of an A.P. is 49 and the sum of its first 17 terms is 289. Find the sum of first n terms of an A.P.****Ans.** S7 = 49, S17 = 289

**Question**. In an AP prove S_{12} = 3 (S_{8} – S_{4}) where S_{n} represent the sum of first n terms of an A.P.**Ans.**

**Question. Ramkali required ₹ 2500 after 12 weeks to send her daughter to school. She saved ₹ 100 in the first week and increased her weekly savings by ₹ 20 every week. Find wheather she will be able to send her daughter to school after 12 weeks.****Ans.** a = 100, d = 20, n = 12

S_{12} = 12/2 [200 + 220] = 6 × 420

= 2520 > 2500

∴ Ram kali will be able to send her daughter to school after 12 weeks.