MCQ Questions | Class 10 Maths Chapter 5 | Arithmetic Progressions with Answers

MCQ | Arithmetic Progressions Class 10 | Sat A

Check the below NCERT MCQ Questions for Class 10 Maths Chapter 5 Arithmetic Progressions with Answers Pdf free download. MCQ Questions for Class 10 Maths with Answers were prepared based on the latest exam pattern. We have Provided Arithmetic Progressions Class 10 Maths MCQs Questions with Answers to help students understand the concept very well.

Objective Question | NCERT Maths Class 10

NCERT Solutions Class 10 Maths
Book:	National Council of Educational Research and Training (NCERT)
Board:	Central Board of Secondary Education (CBSE)
Class:	10th
Subject:	Maths
Chapter:	5
Chapters Name:	Arithmetic Progressions
Medium:	English

Ncert Solutions for Class 10 Maths | Objective Type Questions

You can refer to NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions to revise the concepts in the syllabus effectively and improve your chances of securing high marks in your board exams.

Arithmetic Progressions | Class 10 Maths | NCERT Solutions

Q1.	The n^th term of an A.P. is given by a_n = 3 + 4n. The common difference is
	A. 7
	B. 3
	C. 4
	D. 1

Ans: 4
Explaination:Reason: We have an = 3 + 4n
∴ a_n+1 = 3 + 4(n + 1) = 7 + 4n
∴ d = a_n+1 – a_n
= (7 + 4n) – (3 + 4n)
= 7 – 3
= 4

Q2.	If p, q, r and s are in A.P. then r – q is
	A. s – p
	B. s – q
	C. s – r
	D. none of these

Ans: s – r
Explaination:Reason: Since p, q, r, s are in A.P.
∴ (q – p) = (r – q) = (s – r) = d (common difference)

Q3.	If the sum of three numbers in an A.P. is 9 and their product is 24, then numbers are
	A. 2, 4, 6
	B. 1, 5, 3
	C. 2, 8, 4
	D. 2, 3, 4

Ans: 2, 3, 4
Explaination:Reason: Let three numbers be a – d, a, a + d
∴ a – d +a + a + d = 9
⇒ 3a = 9
⇒ a = 3
Also (a – d) . a . (a + d) = 24
⇒ (3 -d) .3(3 + d) = 24
⇒ 9 – d² = 8
⇒ d² = 9 – 8 = 1
∴ d = ± 1
Hence numbers are 2, 3, 4 or 4, 3, 2

Q4.	The (n – 1)^th term of an A.P. is given by 7,12,17, 22,… is
	A. 5n + 2
	B. 5n + 3
	C. 5n – 5
	D. 5n – 3

Ans: 5n – 3
Explaination:Reason: Here a = 7, d = 12-7 = 5
∴ a_n-1 = a + [(n – 1) – l]d = 7 + [(n – 1) -1] (5) = 7 + (n – 2)5 = 7 + 5n – 10 = 5M – 3

Q5.	The n^th term of an A.P. 5, 2, -1, -4, -7 … is
	A. 2n + 5
	B. 2n – 5
	C. 8 – 3n
	D. 3n – 8

Ans: 8 – 3n
Explaination:Reason: Here a = 5, d = 2 – 5 = -3
a_n = a + (n – 1)d = 5 + (n – 1) (-3) = 5 – 3n + 3 = 8 – 3n

Q6.	The 10^th term from the end of the A.P. -5, -10, -15,…, -1000 is
	A. -955
	B. -945
	C. -950
	D. -965

Ans: -955
Explaination:Reason: Here l = -1000, d = -10 – (-5) = -10 + 5 = – 5
∴ 10^th term from the end = l – (n – 1 )d = -1000 – (10 – 1) (-5) = -1000 + 45 = -955

Q7.	Find the sum of 12 terms of an A.P. whose nth term is given by a_n = 3n + 4
	A. 262
	B. 272
	C. 282
	D. 292

Ans: 262
Explaination:Reason: Here a_n = 3n + 4
∴ a₁ = 7, a₂ – 10, a₃ = 13
∴ a= 7, d = 10 – 7 = 3
∴ S₁₂ = \(\frac{12}{2}\)[2 × 7 + (12 – 1) ×3] = 6[14 + 33] = 6 × 47 = 282

Q8.	The sum of all two digit odd numbers is
	A. 2575
	B. 2475
	C. 2524
	D. 2425

Ans: 2475
Explaination:Reason: All two digit odd numbers are 11,13,15,… 99, which are in A.P.
Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even
∴ Sum = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) × 110 = 45 × 55 = 2475

Q9.	The (n – 1)^th term of an A.P. is given by 7,12,17, 22,… is
	A. 5n + 2
	B. 5n + 3
	C. 5n – 5
	D. 5n – 3

Ans: 5n – 3

Q10.	The number of two digit numbers divisible by 5 is
	A. 19
	B. 18
	C. 16
	D. 17

Ans: 18

Q11.	What is the sum of the first 50 multiples of 3?
	A. 3255
	B. 3825
	C. 4325
	D. 4455

Ans: 3825

Q12.	If 7th and 13th terms of an A.P. be 34 and 64, respectively, then it's 18th term is :
	A. 87
	B. 88
	C. 89
	D. 90

Ans: 89

Q13.	The weights of 11 students selected for a team are noted in ascending order and are in A. P. The lowest value is 45 Kg, and the middle value is 55 Kg. What is the difference between the two values placed consecutively ?
	A. 4
	B. 2
	C. 6
	D. 3

Ans: 2

Q14.	Find the sum of 12 terms of an A.P. whose nth term is given by a_n = 3n + 4
	A. 262
	B. 272
	C. 282
	D. 292

Ans: 262

Q15.	The sum of first n odd natural numbers is
	A. 2n²
	B. 2n + 1
	C. 2n – 1
	D. n²

Ans: n²
Explaination:Reason: Required Sum = 1 + 3 + 5 + … + upto n terms.
Here a = 1, d = 3 – 1 = 2
Sum = \(\frac{n}{2}\)[2 × 1 + (n – 1) × 2] = \(\frac{n}{2}\)[2 + 2n – 2] = \(\frac{n}{2}\) × 2n = n²Reason: All two digit odd numbers are 11,13,15,… 99, which are in A.P.
Since there are 90 two digit numbers of which 45 numbers are odd and 45 numbers are even
∴ Sum = \(\frac{45}{2}\)[11 + 99] = \(\frac{45}{2}\) × 110 = 45 × 55 = 2475

Q16.	The number of multiples lie between n and n² which are divisible by n is
	A. n + 1
	B. n
	C. n – 1
	D. n – 2

Ans: n – 2
Explaination:Reason: Multiples of n from 1 to n² are n × 1, n × 2, n × 3, …, m× n
∴ There are n numbers
Thus, the number of mutiples of n which lie between n and n² is (n – 2) leaving first and last in the given list: Total numbers are (n – 2).

Q17.	If a, b, c, d, e are in A.P., then the value of a – 4b + 6c – 4d + e is
	A. 0
	B. 1
	C. -1.
	D. 2

Ans:
Explaination:Reason: Let common difference of A.P. be x
∴ b = a + x, c = a + 2x, d = a + 3x and e = a + 4x
Given equation n-4b + 6c-4d + c
= a – 4(a + x) + 6(A + 2r) – 4(n + 3x) + (o + 4.v)
= a – 4a – 4x + 6a + 12x – 4a – 12x + a + 4x = 8a – 8a + 16x – 16x = 0

Q18.	n^th term of the sequence a, a + d, a + 2d,… is
	A. a + nd
	B. a – (n – 1)d
	C. a + (n – 1)d
	D. n + nd

Ans: a + nd
Explaination:Reason: an = a + (n – 1)d

Q19.	The 10th term from the end of the A.P. 4, 9,14, …, 254 is
	A. 209
	B. 205
	C. 214
	D. 213

Ans: 29
Explaination:Reason: Here l – 254, d = 9-4 = 5
∴ 10^th term from the end = l – (10 – 1 )d = 254 -9d = 254 = 9(5) = 254 – 45 = 209

Q20.	If 2x, x + 10, 3x + 2 are in A.P., then x is equal to
	A. 0
	B. 2
	C. 4
	D. 6

Ans: 6
Explaination:Reason: Since 2x, x + 10 and 3x + 2 are in A.P.
∴ 2(x + 10) = 2x + (3x + 2)
⇒ 2x + 20 – 5x + 2
⇒ 2x – 5x = 2 – 20
⇒ 3x = 18
⇒ x = 6

CBSE Solutions | Ncert Solutions Maths Class 10

NCERT Solutions | 10th Maths Guide

NCERT | Class 10 Maths Solution

Ncert Solutions for Class 10 Maths

MCQ Questions | Class 10 Maths Chapter 5

MCQ Questions | Class 10 Maths Chapter 5 | Arithmetic Progressions with Answers

MCQ | Arithmetic Progressions Class 10 | Sat A

Objective Question | NCERT Maths Class 10

Ncert Solutions for Class 10 Maths | Objective Type Questions

Arithmetic Progressions | Class 10 Maths | NCERT Solutions

CBSE Solutions | Ncert Solutions Maths Class 10

NCERT Solutions | 10th Maths Guide

NCERT | Class 10 Maths Solution

Ncert Solutions for Class 10 Maths

Ncert Solutions for Class 10 Maths Chapter 1

Ncert Solutions for Class 10 Maths Chapter 2

Ncert Solutions for Class 10 Maths Chapter 3

Ncert Solutions for Class 10 Maths Chapter 4

Ncert Solutions for Class 10 Maths Chapter 5

Ncert Solutions for Class 10 Maths Chapter 6

Ncert Solutions for Class 10 Maths Chapter 7

Ncert Solutions for Class 10 Maths Chapter 8

Ncert Solutions for Class 10 Maths Chapter 9

Ncert Solutions for Class 10 Maths Chapter 10

Ncert Solutions for Class 10 Maths Chapter 11

Ncert Solutions for Class 10 Maths Chapter 12

Ncert Solutions for Class 10 Maths Chapter 13

Ncert Solutions for Class 10 Maths Chapter 14

Ncert Solutions for Class 10 Maths Chapter 15

NCERT SOLUTIONS

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