CBSE Class 10 Mathematics Arithmetic Progressions MCQs Set 23

Multiple Choice Questions (MCQs) for Class 10 Mathematics: Chapter 05 Arithmetic Progressions

Explore reliable objective questions for Chapter 05 Arithmetic Progressions tailored for Class 10 learners. Utilizing these Mathematics multiple-choice formats ensures thorough preparation and strengthens problem-solving speed for upcoming school assessments.

Practice Chapter 05 Arithmetic Progressions MCQs for Class 10 Mathematics

View or download the dedicated Chapter 05 Arithmetic Progressions MCQ resource below. Practicing these 50 objective questions regularly builds familiarity with standard exam patterns and helps secure higher marks in final Mathematics evaluations.

Multiple Choice Questions

Question 1: The 5th and 13th terms of an A.P. are 5 and −3 respectively. Its first term is
(a) 1
(b) −9
(c) 9
(d) 2
Show Answer & Explanation

Answer: (c) 9

Explanation:
1. Subtracting, 8d = −8, so d = −1. Then a = 5 − 4(−1) = 9.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 2: Which term of the A.P. 64, 60, 56, 52, ... is zero?
(a) 16th
(b) 17th
(c) 18th
(d) 15th
Show Answer & Explanation

Answer: (b) 17th

Explanation:
1. 0 = 64 + (n − 1)(−4) gives n − 1 = 16, so n = 17.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 3: The sum of the series 1, 3, 5, 7, 9, ... up to 20 terms is
(a) 420
(b) 210
(c) 400
(d) 380
Show Answer & Explanation

Answer: (c) 400

Explanation:
1. The sum of the first n odd numbers is n² = 400.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 4: The sum of the series 5 + 13 + 21 + ... + 181 is
(a) 2139
(b) 2476
(c) 2219
(d) 2337
Show Answer & Explanation

Answer: (a) 2139

Explanation:
1. d = 8 and 181 = 5 + (n − 1)8 gives n = 23. Sum = 23/2 × (5 + 181) = 23 × 93 = 2139.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 5: The sum of all positive multiples of 5 less than 100 is
(a) 1230
(b) 760
(c) 950
(d) 875
Show Answer & Explanation

Answer: (c) 950

Explanation:
1. The multiples 5, 10, ..., 95 are 19 terms. Sum = 19/2 × (5 + 95) = 950.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 6: The sum of all even natural numbers less than 100 is
(a) 2500
(b) 2272
(c) 2450
(d) 2352
Show Answer & Explanation

Answer: (c) 2450

Explanation:
1. The numbers 2, 4, ..., 98 are 49 terms. Sum = 49/2 × (2 + 98) = 2450.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 7: If 4, A₁, A₂, A₃, 28 are in A.P., then A₃ equals
(a) 23
(b) 19
(c) 24
(d) 22
Show Answer & Explanation

Answer: (d) 22

Explanation:
1. 28 = 4 + 4d gives d = 6. A₃ = 4 + 3(6) = 22.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 8: The number of terms of the A.P. 3, 6, 9, 12, ... that must be taken to make the sum 108 is
(a) 6
(b) 8
(c) 7
(d) 36
Show Answer & Explanation

Answer: (b) 8

Explanation:
1. S_n = n/2 [6 + (n − 1)3] = 108 gives 3n² + 3n = 216, i.e. n² + n − 72 = 0 = (n + 9)(n − 8). So n = 8.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 9: The 6th and 8th terms of an A.P. are 12 and 22. Its 2nd term is
(a) −8
(b) 8
(c) 6
(d) −3
Show Answer & Explanation

Answer: (a) −8

Explanation:
1. Subtracting, 2d = 10, so d = 5 and a = 12 − 25 = −13. a₂ = −13 + 5 = −8.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 10: The sum of the first n terms of an A.P. is 3n²/2 + 5n/2. Its 25th term is
(a) 76
(b) 924
(c) 1924
(d) 74
Show Answer & Explanation

Answer: (a) 76

Explanation:
1. a₂₅ = S₂₅ − S₂₄ = (3(625)/2 + 125/2) − (3(576)/2 + 120/2) = 1000 − 924 = 76.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 11: The third term of an A.P. is 7 and the seventh term exceeds three times the third term by 2. The first term, common difference and the sum of first 20 terms are
(a) −1, 4 and 740
(b) 1, 3 and 610
(c) −1, 4 and 760
(d) −1, 3 and 530
Show Answer & Explanation

Answer: (a) −1, 4 and 740

Explanation:
1. a + 2d = 7 and a + 6d = 3(7) + 2 = 23. Subtracting, 4d = 16, so d = 4 and a = −1. S₂₀ = 10[−2 + 19(4)] = 740.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 12: The first term of an A.P. is 2 and the last term is 50. The sum of all the terms is 442. The common difference is
(a) 3
(b) 2
(c) 4
(d) 5
Show Answer & Explanation

Answer: (a) 3

Explanation:
1. S = n/2 (2 + 50) = 26n = 442 gives n = 17. Then 50 = 2 + 16d gives d = 3.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 13: The 12th term of an A.P. is −13 and the sum of its first four terms is 24. The sum of its first 10 terms is
(a) −10
(b) 0
(c) 10
(d) −5
Show Answer & Explanation

Answer: (b) 0

Explanation:
1. a + 11d = −13 and 4a + 6d = 24, i.e. 2a + 3d = 12. From the first, a = −13 − 11d, so −26 − 22d + 3d = 12 and d = −2, a = 9. S₁₀ = 5[18 + 9(−2)] = 0.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 14: The 5th and 12th terms of an A.P. are 30 and 65. The sum of its first 20 terms is
(a) 1150
(b) 1050
(c) 1250
(d) 1000
Show Answer & Explanation

Answer: (a) 1150

Explanation:
1. 7d = 35, so d = 5 and a = 30 − 20 = 10. S₂₀ = 10[20 + 19(5)] = 1150.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 15: A factory increases its production of TV sets by a fixed number every year. It produced 8000 sets in the 6th year and 11300 in the 9th year. The production in the 8th year was
(a) 9100
(b) 11300
(c) 10200
(d) 9000
Show Answer & Explanation

Answer: (c) 10200

Explanation:
1. d = 1100. 8th year = 8000 + 2(1100) = 10200.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 16: Rs. 2800 is used to award four prizes, each prize after the first being Rs. 200 less than the preceding one. The first prize is
(a) Rs. 1000
(b) Rs. 800
(c) Rs. 1200
(d) Rs. 700
Show Answer & Explanation

Answer: (a) Rs. 1000

Explanation:
1. The prizes are a, a − 200, a − 400, a − 600 with sum 4a − 1200 = 2800, so a = 1000. The prizes are Rs. 1000, 800, 600, 400.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 17: The angles of a quadrilateral are in A.P. with common difference 10°. The angles are
(a) 75°, 85°, 95°, 105°
(b) 70°, 80°, 90°, 100°
(c) 80°, 90°, 100°, 110°
(d) 65°, 75°, 85°, 95°
Show Answer & Explanation

Answer: (a) 75°, 85°, 95°, 105°

Explanation:
1. 4a + 6(10) = 360 gives a = 75. The angles are 75°, 85°, 95°, 105°.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 18: The sum of three terms of an A.P. is 21 and the product of the first and third terms exceeds the second term by 6. The three terms are
(a) 3, 7, 11
(b) 2, 7, 12
(c) 5, 7, 9
(d) 1, 7, 13
Show Answer & Explanation

Answer: (d) 1, 7, 13

Explanation:
1. Let the terms be 7 − d, 7, 7 + d (sum 21). (7 − d)(7 + d) − 7 = 6 gives 49 − d² = 13, d² = 36, d = ±6. The terms are 1, 7, 13 (or 13, 7, 1).

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 19: Four numbers in A.P. have sum 50 and the greatest is 4 times the least. The numbers are
(a) 8, 12, 16, 20
(b) 10, 15, 20, 25
(c) 5, 10, 15, 20
(d) 4, 12, 20, 28
Show Answer & Explanation

Answer: (c) 5, 10, 15, 20

Explanation:
1. 4a + 6d = 50 and a + 3d = 4a, so d = a. Then 10a = 50, a = 5. The numbers are 5, 10, 15, 20.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 20: The sum of all odd integers between 1 and 1000 which are divisible by 3 is
(a) 83500
(b) 83667
(c) 84500
(d) 83000
Show Answer & Explanation

Answer: (b) 83667

Explanation:
1. The numbers 3, 9, 15, ..., 999 form an A.P. with d = 6 and 167 terms. Sum = 167/2 × (3 + 999) = 167 × 501 = 83667.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 21: How many numbers between 103 and 750 are divisible by 6?
(a) 108
(b) 107
(c) 125
(d) 113
Show Answer & Explanation

Answer: (b) 107

Explanation:
1. The numbers are 108, 114, ..., 744. 744 = 108 + (n − 1)6 gives n − 1 = 106, so n = 107. (Equivalently 744/6 − 18 + 1 = 107.)

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Question 22: How many two-digit numbers are divisible by 7?
(a) 12
(b) 14
(c) 15
(d) 13
Show Answer & Explanation

Answer: (d) 13

Explanation:
1. The numbers are 14, 21, ..., 98. 98 = 14 + (n − 1)7 gives n = 13.

Teacher's Note:
1. Write the standard formulas a_n = a + (n − 1)d and S_n = n/2 [2a + (n − 1)d], form equations from the given data and solve them.
2. Check by listing the first few terms.

Free study material for Mathematics

Chapter 05 Arithmetic Progressions Objective Questions & Solutions for Class 10 Mathematics

Chapter MCQs with Answers for Class 10 Mathematics

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Expert Practice Material for Class 10 Mathematics

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FAQs

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