# Assignments Class 10 Mathematics Arithmetic Progression

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

Question. Find 5th term of an A.P. whose nth term is 3n – 5
Ans. an = 3n – 5, a5 = 10

Question. Write the common difference of an A.P. whose nth term is an = 3n + 7
Ans. a1 = 3 + 7 = 10, a2 = 6 + 7 = 13, d = 3

Question. Find the sum of first 10 even numbers.
Ans. Sn = 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 a8 – a4 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 10th 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. nth term of an A.P. 5, 10, 15, 20 …. n terms and nth 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, nth term = k times nth term of original A.P. of new A.P.

Question. Difference of mth and nth 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, ∵ Sn = (n(n+1))/2 = (20×21)/2 = 210

Question. Which term of the A.P. 5, 15, 25, ……. will be 130 more than its 31st term?
Ans. Let an = 130 + a31
Solve to get n = 44

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 10th term is equal to 20 times of 20th term of an A.P. Find its 30th term.
Ans. ATQ 10 a10 = 20 a20
⇒ a10 = 2a20
a + 9d = 2a + 38d
a = – 29d               …(1)
a30 = a + 29d
Substitute a from (1)
∴ Answer is a30 = 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.
a2 – a1 = a3 – a2 = 6
∴ A.P., an = 96 ⇒ n = 15

Question. Which term of the A.P. : 121, 117, 113 … is the first negative terms ?
Ans. Let an < 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 Sn, where an = 5n – 1. Hence find the sum of the first 20 terms.
Ans.

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
an = 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 an = a + (n – 1) d , then use Sn = n/2 [2a + (n – 1) d]
∴ Answer is Sn = 17885. (n = 70)

Question. For what value of n, are the nth term of two A.P’s 63, 65, 67 ……… and 3, 10, 17 ….. are equal ?
Ans. 63, 65, 67, …..
an = 63 + (n – 1) 2 = 61 + 2n
3, 10, 17, ….
an = 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 qth term is p, prove that its nth term is (p + q – n).
Ans. ap = q, aq = p
Solve to get a and d, a = q + p – 1, d = – 1
an = p + q – n

Question. The 17th term of an A.P. is 5 more than twice its 8th term. If the 11th term of the A.P. is 43, then find the nth term of the A.P.
Ans. ATQ,
a17 = 5 + 2 × a8
a + 16d = 5 + 2a + 14 d
a – 2d = – 5                       …(1)
a11 = a + 10d = 43            ….(2)
Solving (1) & (2), we get
a = 3, d = 4
∴ an = 4n – 1

Question. The sum of 5th and 9th terms of an A.P. is 30. If its 25th term is three times its 8th term, find the A.P.
Ans. ATQ a5 + a9 = 30
a25 = 3 a8
Solve to get a = 3, d = 2
A.P. 3, 5, 7, 9, …

Question. If Sn, the sum of first n terms of an A.P. is given by Sn = 3n2 – 4n, find the nth term.
Ans. Sn = 3n2 – 4n
an = Sn – Sn–1
= (3n2 – 4n) – [3(n – 1)2 – 4(n – 1)]
= (3n2 – 4n) – [3n2 + 3 – 6n – 4n + 4]
= – [7 – 6n]
an = 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 20th term.
Ans.

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

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 S20 = 1/3 (S40 – S20), a = 1
Use Sn = n/2 [2a + (n – 1) d] and a = 1 to find d, d = 2
then find S30

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 4th term is 18 and the difference of 9th term from the 15th term is 30.
Ans. ATQ a4 = 18     …(1), a15 – a9 = 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 S12 = 3 (S8 – S4) where Sn 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
S12 = 12/2 [200 + 220] = 6 × 420
= 2520 > 2500
∴ Ram kali will be able to send her daughter to school after 12 weeks.