CBSE Class 10 Mathematics Arithmetic Progression MCQs Set 16

Practice MCQs for Class 10 Mathematics Chapter 05 Arithmetic Progression

Review structured MCQ sets for Class 10 Mathematics Chapter 05 Arithmetic Progression. Built according to official CBSE guidelines, these downloadable questions support daily revision and core concept reinforcement.

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Navigate directly to the 50 objective questions for Chapter 05 Arithmetic Progression using the digital viewer below. Each practice set includes verified answer keys, allowing students to instantly cross-check their work and identify areas requiring further revision.

Multiple Choice Questions

Question: In an A.P., if the first term \(a = -16\) and the common difference \(d = -2\), then the sum of the first 10 terms is
(a) \(-200\)
(b) \(-70\)
(c) \(-250\)
(d) \(250\)
Show Answer & Explanation

Answer: (c) \(-250\)

Explanation:
1. \(S_{10} = \frac{10}{2}[2(-16) + 9(-2)] = 5(-50) = -250\).

Question: If the sum of the first \(n\) terms of an A.P. is \(3n^2 + n\) and its common difference is 6, then its first term is
(a) 2
(b) 3
(c) 1
(d) 4
Show Answer & Explanation

Answer: (d) 4

Explanation:
1. \(S_n = \frac{n}{2}[2a + 6(n - 1)] = 3n^2 + n\), which gives \(2an = 8n\).
2. So \(a = 4\).

Question: The \(n\)th term of the AP \(a, 3a, 5a, \ldots\) is
(a) \(na\)
(b) \((2n - 1)a\)
(c) \((2n + 1)a\)
(d) \(2na\)
Show Answer & Explanation

Answer: (b) \((2n - 1)a\)

Explanation:
1. \(d = 2a\), so \(a_n = a + (n - 1)2a = (2n - 1)a\).

Question: The common difference of the AP \(\frac{1}{p}, \frac{1 - p}{p}, \frac{1 - 2p}{p}, \ldots\) is
(a) \(1\)
(b) \(\frac{1}{p}\)
(c) \(-1\)
(d) \(-\frac{1}{p}\)
Show Answer & Explanation

Answer: (c) \(-1\)

Explanation:
1. \(d = \frac{1 - p}{p} - \frac{1}{p} = \frac{-p}{p} = -1\).

Question: The value of \(x\) for which \(2x,\ (x + 10)\) and \((3x + 2)\) are three consecutive terms of an AP is
(a) 6
(b) -6
(c) 18
(d) -18
Show Answer & Explanation

Answer: (a) 6

Explanation:
1. \((x + 10) - 2x = (3x + 2) - (x + 10)\), so \(-x + 10 = 2x - 8\).
2. \(3x = 18\), so \(x = 6\).

Question: The 4th term from the end of the AP \(-11, -8, -5, \ldots, 49\) is
(a) 37
(b) 40
(c) 43
(d) 58
Show Answer & Explanation

Answer: (b) 40

Explanation:
1. \(d = 3\), last term \(l = 49\).
2. 4th term from the end \(= 49 - 3 \times 3 = 40\).

Question: The famous mathematician associated with finding the sum of the first 100 natural numbers is
(a) Pythagoras
(b) Newton
(c) Gauss
(d) Euclid
Show Answer & Explanation

Answer: (c) Gauss

Explanation:
1. Gauss found that \(1 + 2 + \cdots + 100 = 5050\).

Question: If the first term of an AP is \(-5\) and the common difference is 2, then the sum of the first 6 terms is
(a) 0
(b) 5
(c) 6
(d) 15
Show Answer & Explanation

Answer: (a) 0

Explanation:
1. \(S_6 = 3[2(-5) + 5(2)] = 3(0) = 0\).

Question: The sum of the first 16 terms of the AP \(10, 6, 2, \ldots\) is
(a) -320
(b) 320
(c) -352
(d) -400
Show Answer & Explanation

Answer: (a) -320

Explanation:
1. \(d = -4\). \(S_{16} = 8[20 + 15(-4)] = 8(-40) = -320\).

Question: In an AP, if \(a = 1\), \(a_n = 20\) and \(S_n = 399\), then \(n\) is equal to
(a) 19
(b) 21
(c) 38
(d) 42
Show Answer & Explanation

Answer: (c) 38

Explanation:
1. \(S_n = \frac{n}{2}(1 + 20) = 399\), so \(n = \frac{798}{21} = 38\).

Question: The sum of the first five multiples of 3 is
(a) 45
(b) 55
(c) 65
(d) 75
Show Answer & Explanation

Answer: (a) 45

Explanation:
1. The multiples are 3, 6, 9, 12, 15. \(S_5 = \frac{5}{2}(3 + 15) = 45\).

Question: The number of two-digit numbers which are divisible by 3 is
(a) 33
(b) 31
(c) 30
(d) 29
Show Answer & Explanation

Answer: (c) 30

Explanation:
1. The numbers are 12, 15, \ldots, 99 with \(d = 3\). \(99 = 12 + (n - 1)3\), so \(n = 30\).

Question: The first four terms of an AP whose first term is \(-2\) and common difference is \(-2\) are
(a) \(-2, 0, 2, 4\)
(b) \(-2, 4, -8, 16\)
(c) \(-2, -4, -6, -8\)
(d) \(-2, -4, -8, -16\)
Show Answer & Explanation

Answer: (c) \(-2, -4, -6, -8\)

Explanation:
1. Terms: \(-2\), \(-4\), \(-6\), \(-8\).

Question: The 21st term of an AP whose first two terms are \(-3\) and \(4\) is
(a) 17
(b) 137
(c) 143
(d) -143
Show Answer & Explanation

Answer: (b) 137

Explanation:
1. \(d = 7\). \(a_{21} = -3 + 20 \times 7 = 137\).

Question: If the 2nd term of an AP is 13 and the 5th term is 25, what is its 7th term?
(a) 30
(b) 33
(c) 37
(d) 38
Show Answer & Explanation

Answer: (b) 33

Explanation:
1. \(a + d = 13\) and \(a + 4d = 25\), so \(d = 4\), \(a = 9\).
2. \(a_7 = 9 + 6 \times 4 = 33\).

Question: An AP starts with a positive fraction and every alternate term is an integer. If the sum of the first 11 terms is 33, then the fourth term is
(a) 2
(b) 3
(c) 5
(d) 6
Show Answer & Explanation

Answer: (a) 2

Explanation:
1. \(S_{11} = 11(a + 5d) = 33\), so \(a_6 = 3\).
2. Alternate terms are integers, so \(a_4 = 2\).

Question: If the sum of the first \(2n\) terms of \(2, 5, 8, \ldots\) equals the sum of the first \(n\) terms of \(57, 59, 61, \ldots\), then \(n\) is equal to
(a) 10
(b) 12
(c) 11
(d) 13
Show Answer & Explanation

Answer: (c) 11

Explanation:
1. \(\frac{2n}{2}[4 + (2n - 1)3] = \frac{n}{2}[114 + 2(n - 1)]\), which gives \(10n = 110\).
2. So \(n = 11\).

Question: In an AP, if \(d = -4\), \(n = 7\) and \(a_n = 4\), then \(a\) is equal to
(a) 6
(b) 7
(c) 20
(d) 28
Show Answer & Explanation

Answer: (d) 28

Explanation:
1. \(4 = a + 6(-4)\), so \(a = 28\).

Question: The sum of five distinct positive integers in AP is 10020 with a positive common difference. The smallest possible value of the last term is
(a) 2002
(b) 2004
(c) 2006
(d) 2007
Show Answer & Explanation

Answer: (c) 2006

Explanation:
1. Terms: \(a - 2d, a - d, a, a + d, a + 2d\). \(5a = 10020\), so \(a = 2004\).
2. Smallest \(d = 1\), so the last term is \(2004 + 2 = 2006\).

Question:
Assertion (A): If the \(n\)th term of an AP is \(7 - 4n\), then its common difference is \(-4\).
Reason (R): The common difference of an AP is given by \(d = a_{n+1} - a_n\).
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \(a_{n+1} - a_n = -4\), so the assertion is true. The reason is the definition of d and explains it.

Question:
Assertion (A): The common difference of the AP \(-5, -1, 3, 7, \ldots\) is 4.
Reason (R): The common difference of the AP \(a, a + d, a + 2d, \ldots\) is \(d = a_2 - a_1\).
(a) Both A and R are true, and R is the correct explanation of A.
(b) Both A and R are true, but R is not the correct explanation of A.
(c) A is true, but R is false.
(d) A is false, but R is true.
Show Answer & Explanation

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \(d = -1 - (-5) = 4\), so the assertion is true. The reason is the definition of d and explains the value 4.

Teacher's Note:
1. The source marks option (c). That is wrong, because the reason is true and explains the assertion. Corrected to option (a).

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Download Chapter MCQs: Class 10 Mathematics Chapter 05 Arithmetic Progression

Download Multiple Choice Questions: Chapter 05 Arithmetic Progression (Class 10 Mathematics)

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FAQs

Where can I access latest CBSE Class 10 Mathematics Arithmetic Progression MCQs Set 16?

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Yes, our CBSE Class 10 Mathematics Arithmetic Progression MCQs Set 16 include the latest type of questions, such as Assertion-Reasoning and Case-based MCQs. 50% of the CBSE paper is now competency-based.

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