MCQs on Chapter 5: Arithmetic Progressions for Class 10 Maths

Arithmetic Progressions Class 10 MCQs are available in this Maths article. These are multiple-choice questions that enable students to practise important concepts from Chapter 5 of the CBSE Maths syllabus in an exam-oriented manner. MCQs with answers and detailed solutions are provided for arithmetic progression, common difference, nth term, sum of n terms, and application of formulas to enhance conceptual understanding and improve problem-solving skills. Solving the MCQs on arithmetic progressions helps the students to improve the accuracy, understand the formulas better and gain confidence for the board exams.

MCQS on Chapter 5: Arithmetic Progressions for Class 10 With Answers

Question 1: Which of the following is an Arithmetic Progression?

(a) 1, 1, 2, 3, 5, 8, …

(b) 1, 4, 9, 16, 25, …

(c) 3, 7, 11, 15, 19, …

(d) 1, 2, 4, 8, 16, …

Answer: (c) 3, 7, 11, 15, 19, …

Explanation: Check the differences between consecutive terms in each option:

(a) 1−1=0, 2−1=1, 3−2=1; differences are not constant. Not an AP.

(b) 4−1=3, 9−4=5, 16−9=7; not constant. Not an AP (perfect squares).

(c) 7−3=4, 11−7=4, 15−11=4; constant difference of 4. This is an AP.

(d) 2−1=1, 4−2=2, 8−4=4; not constant. Not an AP

Question 2: For what value of k are 2k, k+10 and 3k+2 in arithmetic progression?

(a) k = 4

(b) k = 6

(c) k = 8

(d) k = 2

Answer: (b) k = 6

Explanation: Three terms a, b, c are in AP if and only if 2b = a + c.

So: 2(k + 10) = 2k + (3k + 2)

⇒ 2k + 20 = 5k + 2

⇒ 18 = 3k ⇒ k = 6

Question 3: The common difference of the AP: −5, −5, −5, −5, … is:

(a) −5

(b) 5

(c) 0

(d) Not an AP

Answer: (c) 0

Explanation: d = (−5) − (−5) = 0. A constant sequence is a valid AP.

Question 4: The 30th term of the AP: 10, 7, 4, … is:

(a) 97

(b) −67

(c) −77

(d) −87

Answer:  (c) −77

Explanation: a = 10, d = 7 − 10 = −3

a₃₀ = 10 + (30 − 1)(−3) = 10 + 29 × (−3) = 10 − 87 = −77

⇒ a₃₀ = −77. 

Question 5: Which term of the AP: 3, 8, 13, 18, … is 78?

(a) 12th

(b) 13th

(c) 15th

(d) 16th

Answer: (d) 16th

Explanation: a = 3, d = 5. Set aₙ = 78:

78 = 3 + (n − 1) × 5

75 = (n − 1) × 5 ⇒ n − 1 = 15 ⇒ n = 16

78 is the 16th term.

Question 6: Is 301 a term of the AP: 5, 11, 17, 23, …?

(a) Yes, it is the 50th term

(b) Yes, it is the 49th term

(c) No, 301 does not belong to this AP

(d) Yes, it is the 51st term

Answer:  (c) No, 301 does not belong to this AP

Explanation: a = 5, d = 6. Set aₙ = 301:

301 = 5 + (n − 1) × 6 ⇒ 296 = (n − 1) × 6 ⇒ n − 1 = 296/6 ≈ 49.33

Since n is not a whole number, 301 does not belong to this AP.

Question 7: The sum of the AP: 7, 13, 19, …, 205 is:

(a) 3432

(b) 3528

(c) 3604

(d) 2100

Answer:  (c) 3604

Explanation: a = 7, d = 6, l = 205

First find n: 205 = 7 + (n−1)×6 ⇒ 198 = (n−1)×6 ⇒ n = 34

Now use alternate formula: S₃₄ = 34/2 × (7 + 205) = 17 × 212 = 3604

Question 8: How many terms of the AP: 9, 17, 25, … must be taken to give a sum of 636?

(a) 12

(b) 11

(c) 10

(d) 13

Answer:  (a) 12

Explanation: a = 9, d = 8. Set Sₙ = 636:

n/2 [2(9) + (n−1)(8)] = 636

n/2 [18 + 8n − 8] = 636 ⇒ n(8n + 10) = 1272 ⇒ 8n² + 10n − 1272 = 0

4n² + 5n − 636 = 0 ⇒ (4n + 53)(n − 12) = 0 ⇒ n = 12 (rejecting negative)

n = 12 terms.

Question 9: If the sum of first p terms of an AP is the same as the sum of its first q terms (p ≠ q), the sum of its first (p + q) terms is:

(a) p + q

(b) pq

(c) 0

(d) 2(p + q)

Answer:  (c) 0

Explanation: Given S_p = S_q

p/2[2a + (p−1)d] = q/2[2a + (q−1)d]

Simplifying: 2a(p − q) + d(p² − p − q² + q) = 0

2a(p−q) + d(p−q)(p+q−1) = 0

Since p ≠ q: 2a + d(p+q−1) = 0

Now S_(p+q) = (p+q)/2 × [2a + (p+q−1)d] = (p+q)/2 × 0 = 0

S_(p+q) = 0.

Question 10: Three numbers are in AP and their sum is 27. If the largest is doubled and the smallest is trebled, the new sum is 57. Find the three numbers.

(a) 7, 9, 12

(b) 6, 9, 12 

(c) 5, 9, 13

(d) 7, 9, 11

Answer:  (b) 6, 9, 12 

Explanation: When three numbers are in AP, take them as (a − d), a, (a + d).

Sum: (a − d) + a + (a + d) = 3a = 27  ⇒ a = 9

Condition: 2(a + d) + a + 3(a − d) = 57

2a + 2d + a + 3a − 3d = 57  ⇒ 6a − d = 57  ⇒ 54 − d = 57  ⇒ d = −3

Numbers: (9 − (−3)), 9, (9 + (−3)) = 12, 9, 6

Question 11: The 4th term from the end of the AP: 11, 8, 5, 2, …, −31 is:

(a) −25

(b) −22

(c) −19

(d) −28

Answer: (b) −22

Explanation: Method: Treat last term as the first term of the reversed AP, with a common difference of −d.

l = −31, d_reversed = −(−3) = 3

4th term from end = l − (4 − 1)d = −31 − (3)(−3) = −31 + 9 = −22 

Question 12: Subba Rao started work in 1995 at an annual salary of ₹5000 and received an increment of ₹200 each year. In which year did his income reach ₹7000?

(a) 2004

(b) 2005

(c) 2006

(d) 2007

Answer: (b) 2005

Explanation: a = 5000, d = 200, aₙ = 7000

7000 = 5000 + (n − 1) × 200

2000 = (n − 1) × 200 ⇒ n − 1 = 10 ⇒ n = 11

He started in 1995 (1st year). The 11th year is 1995 + 10 = 2005.

Question 13: In a school, students thought of planting trees in rows to form a triangle. The first row had 1 tree, the second row 2 trees, the third row 3, and so on. There were 12 rows in all. How many trees were planted?

(a) 66

(b) 72

(c) 78

(d) 60

Answer:  (c) 78

Explanation: AP: 1, 2, 3, …, 12 with a = 1, d = 1, n = 12

S₁₂ = 12/2 × (1 + 12) = 6 × 13 = 78

78 trees were planted.

Question 14: A theatre has 30 seats in the first row, 32 in the second, 34 in the third, and so on. The last row has 100 seats. How many seats does the theatre have in total?

(a) 1950

(b) 2340

(c) 1950

(d) 2160

Answer: (b) 2340

 Explanation: AP: 30, 32, 34, …, 100 with a = 30, d = 2, l = 100

First find n: 100 = 30 + (n−1)×2  ⇒ 70 = (n−1)×2  ⇒ n = 36

Total seats = S₃₆ = 36/2 × (30 + 100) = 18 × 130 = 2,340 seats. 

Question 15: The sum of first 7 terms of an AP is 63, and the sum of its next 7 terms is 161. Find the 28th term.

(a) 57

(b) 60

(c) 57

(d) 65

Answer: (c) 57 

Explanation: S₇ = 63 ⇒ 7/2 [2a + 6d] = 63 ⇒ 2a + 6d = 18 ⇒ a + 3d = 9 … (i)

S₁₄ = 63 + 161 = 224 ⇒ 14/2 [2a + 13d] = 224 ⇒ 2a + 13d = 32 … (ii)

2×(i): 2a + 6d = 18. Subtract from (ii): 7d = 14 ⇒ d = 2, a = 3

a₂₈ = 3 + 27 × 2 = 3 + 54 = 57

a₂₈ = 57. 

Click here to download the free PDF of the MCQs worksheet on Chapter 5 Arithmetic Progressions for Class 10 Maths based on the updated NCERT & CBSE pattern with important multiple-choice questions and answers.

MCQs Worksheet on Chapter 5: Arithmetic Progressions for Class 10