Class 10 Mathematics Study Guide: CBSE Class 10 Mathematics Arithmetic Progressions Important Questions Set 01
Access comprehensive study materials and useful resources for Chapter 05 Arithmetic Progressions using the CBSE Class 10 Mathematics Arithmetic Progressions Important Questions Set 01. Designed to align with the 2026-27 CBSE academic guidelines, these advanced resources help Class 10 Mathematics students reinforce core concepts beyond standard textbooks.
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Questions
Section A: Objective and Very Short Answer (1 mark each)
Q1. Write the fourth term of an AP whose nth term is 3n + 2. [1 mark]
View Answer & Explanation
Answer: a₄ = 3(4) + 2 = 14.
Teacher's Note:
1. Put n = 4 in the formula.
2. No other information is needed.
Q2. Find the first term of an AP whose fifth term is 5 and common difference is −3. [1 mark]
View Answer & Explanation
Answer: a + 4d = 5 gives a = 5 + 12 = 17. The AP is 17, 14, 11, 8, 5, …
Teacher's Note:
1. Use aₙ = a + (n − 1)d.
2. Check by counting down four steps from 17.
Q3. Find the 10th term from the end of the AP 4, 9, 14, …, 254. [1 mark]
View Answer & Explanation
Answer: Here d = 5 and the last term l = 254. The 10th term from the end = l − 9d = 254 − 45 = 209.
Teacher's Note:
1. Use l − (n − 1)d for terms from the end.
2. Alternatively reverse the AP with d = −5.
Q4. Is 0 a term of the AP 40, 37, 34, 31, …? [1 mark]
View Answer & Explanation
Answer: aₙ = 40 − 3(n − 1) = 0 gives n − 1 = 40/3, which is not an integer. So no, 0 is not a term.
Teacher's Note:
1. Set aₙ = 0 and solve for n.
2. n must be a positive integer.
Q5. For what value of x are x + 2, 2x and 2x + 3 three consecutive terms of an AP? [1 mark]
View Answer & Explanation
Answer: 2(2x) = (x + 2) + (2x + 3), so 4x = 3x + 5 and x = 5.
Teacher's Note:
1. The middle term is the average of its neighbours.
2. Equate 2b = a + c.
Q6. Find the sum of the first 24 terms of the AP 5, 8, 11, 14, … [1 mark]
View Answer & Explanation
Answer: S₂₄ = (24/2)[2(5) + 23(3)] = 12 × 79 = 948.
Teacher's Note:
1. Use Sₙ = (n/2)[2a + (n − 1)d].
2. Here a = 5, d = 3.
Q7. Which term of the AP 12, 7, 2, −3, … is −98? [1 mark]
View Answer & Explanation
Answer: 12 − 5(n − 1) = −98, so n − 1 = 22 and n = 23.
Teacher's Note:
1. The common difference is −5.
2. Solve aₙ = −98 for n.
Q8. If the nth term of an AP is 3n + 5, find its common difference. [1 mark]
View Answer & Explanation
Answer: a₁ = 8 and a₂ = 11, so d = 3.
Teacher's Note:
1. Subtract consecutive terms.
2. The coefficient of n gives d directly.
Q9. Write the next term of the AP \( \sqrt{2},\ \sqrt{18},\ \dots \). [1 mark]
View Answer & Explanation
Answer: √18 = 3√2, so d = 3√2 − √2 = 2√2. The next term is 3√2 + 2√2 = 5√2 (that is, √50).
Teacher's Note:
1. Simplify the radicals first.
2. Add the common difference.
Q10. If 4/5, a, 2 are three consecutive terms of an AP, find a. [1 mark]
View Answer & Explanation
Answer: a = (4/5 + 2)/2 = \( \dfrac{7}{5} \).
Teacher's Note:
1. The middle term is the mean of the other two.
2. Add and halve.
Q11. The 10th term of the AP 2, 7, 12, … is [1 mark]
(a) 42
(b) 45
(c) 47
(d) 52
View Answer & Explanation
Answer: (c) 47. a₁₀ = 2 + 9 × 5 = 47.
Teacher's Note:
1. Here a = 2 and d = 5.
2. Use a + 9d.
Q12. If k, 2k − 1 and 2k + 1 are three consecutive terms of an AP, then k is [1 mark]
(a) 2
(b) 3
(c) −3
(d) 5
View Answer & Explanation
Answer: (b) 3. 2(2k − 1) = k + 2k + 1 gives 4k − 2 = 3k + 1, so k = 3.
Teacher's Note:
1. Use 2 × middle = sum of the others.
2. Solve the linear equation.
Section B: Short Answer (2 and 3 marks)
Q13. Find the middle term of the AP 6, 13, 20, …, 216. [2 marks]
View Answer & Explanation
Answer: 216 = 6 + 7(n − 1) gives n = 31. The middle term is the 16th: a₁₆ = 6 + 15 × 7 = 111.
Teacher's Note:
1. Find the number of terms first.
2. The middle of 31 terms is the 16th.
Q14. The 6th term of an AP is −10 and its 10th term is −26. Find the 15th term. [2 marks]
View Answer & Explanation
Answer: a + 5d = −10 and a + 9d = −26, so 4d = −16 and d = −4, a = 10.
a₁₅ = 10 + 14(−4) = −46.
Teacher's Note:
1. Subtract the two equations to find d.
2. Then find a and the required term.
Q15. If the 8th term of an AP is 0, prove that its 38th term is triple its 18th term. [3 marks]
View Answer & Explanation
Answer: a + 7d = 0, so a = −7d.
a₃₈ = a + 37d = 30d and a₁₈ = a + 17d = 10d.
So a₃₈ = 3 × a₁₈. Proved.
Teacher's Note:
1. Replace a by −7d.
2. Compare the two terms.
Q16. The sum of three numbers in AP is 21 and their product is 231. Find the numbers. [3 marks]
View Answer & Explanation
Answer: Let the numbers be a − d, a, a + d. Sum = 3a = 21, so a = 7.
Product: 7(49 − d²) = 231 gives d² = 16, d = ±4.
The numbers are 3, 7, 11 (or 11, 7, 3).
Teacher's Note:
1. Take the terms symmetrically around the middle one.
2. The sum gives the middle term directly.
Q17. Find the sum of the first 25 terms of an AP whose nth term is 7 − 3n. [3 marks]
View Answer & Explanation
Answer: a₁ = 4 and a₂₅ = 7 − 75 = −68.
S₂₅ = (25/2)(4 − 68) = 25 × (−32) = −800.
Teacher's Note:
1. Use Sₙ = (n/2)(first + last).
2. Find both terms from the formula.
Q18. Find the sum of all two-digit odd positive numbers. [3 marks]
View Answer & Explanation
Answer: The numbers 11, 13, …, 99 form an AP with d = 2. 99 = 11 + 2(n − 1) gives n = 45.
S = (45/2)(11 + 99) = 2475.
Teacher's Note:
1. List the first and last terms.
2. Count the terms before summing.
Q19. Find the sum of all three-digit numbers divisible by 11. [3 marks]
View Answer & Explanation
Answer: The numbers 110, 121, …, 990 form an AP with d = 11. 990 = 110 + 11(n − 1) gives n = 81.
S = (81/2)(110 + 990) = 44550.
Teacher's Note:
1. The first three-digit multiple of 11 is 110.
2. The last is 990.
Q20. Find the sum of all three-digit numbers that are multiples of 7. [3 marks]
View Answer & Explanation
Answer: The numbers 105, 112, …, 994 form an AP with d = 7. 994 = 105 + 7(n − 1) gives n = 128.
S = (128/2)(105 + 994) = 64 × 1099 = 70336.
Teacher's Note:
1. The first multiple of 7 above 99 is 105.
2. The last below 1000 is 994.
Q21. The sum of the first 6 terms of an AP is 42, and the ratio of its 10th term to its 30th term is 1 : 3. Find the first term and the 13th term. [3 marks]
View Answer & Explanation
Answer: S₆ = 3(2a + 5d) = 42, so 2a + 5d = 14.
(a + 9d)/(a + 29d) = 1/3 gives 3a + 27d = a + 29d, so a = d. Then 7a = 14, a = 2, d = 2.
a₁₃ = 2 + 12 × 2 = 26. First term 2, 13th term 26.
Teacher's Note:
1. Form two equations in a and d.
2. The ratio condition gives a = d.
Q22. How many terms of the AP 17, 15, 13, … must be added to get the sum 72? Explain the double answer. [3 marks]
View Answer & Explanation
Answer: (n/2)[34 − 2(n − 1)] = 72 gives n(18 − n) = 72, so n² − 18n + 72 = 0 and n = 6 or n = 12.
The 7th to 12th terms are 5, 3, 1, −1, −3, −5, which add up to zero, so the sum stays 72 at n = 12 also.
Teacher's Note:
1. Solve the quadratic in n.
2. Negative terms later cancel the positive ones.
Q23. Is 51 a term of the AP 5, 8, 11, 14, …? [3 marks]
View Answer & Explanation
Answer: 5 + 3(n − 1) = 51 gives n − 1 = 46/3, not an integer. So 51 is not a term.
Teacher's Note:
1. The terms are 5 plus multiples of 3.
2. 51 − 5 = 46 is not a multiple of 3.
Q24. If 2x, x + 10 and 3x + 2 are in AP, find x. [3 marks]
View Answer & Explanation
Answer: 2(x + 10) = 2x + 3x + 2, so 2x + 20 = 5x + 2 and x = 6.
Teacher's Note:
1. The middle term is twice itself on the left.
2. Solve the linear equation.
Q25. The first term of an AP is −7 and the common difference is 5. Find its 18th term and its general term. [3 marks]
View Answer & Explanation
Answer: aₙ = −7 + 5(n − 1) = 5n − 12. a₁₈ = 90 − 12 = 78.
Teacher's Note:
1. Use aₙ = a + (n − 1)d.
2. Simplify to get the general term.
Q26. The sum of the first n terms of an AP is (3n² + 5n)/2. Find its 25th term. [3 marks]
View Answer & Explanation
Answer: S₂₅ = (1875 + 125)/2 = 1000 and S₂₄ = (1728 + 120)/2 = 924.
a₂₅ = S₂₅ − S₂₄ = 76.
Teacher's Note:
1. The nth term equals Sₙ − Sₙ₋₁.
2. Here aₙ = 3n + 1 gives the same value.
Section C: Long Answer (4 marks)
Q27. The sums of the first n, 2n and 3n terms of an AP are S₁, S₂ and S₃. Prove that S₃ = 3(S₂ − S₁). [4 marks]
View Answer & Explanation
Answer: S₁ = (n/2)[2a + (n − 1)d], S₂ = n[2a + (2n − 1)d].
S₂ − S₁ = (n/2)[4a + 2(2n − 1)d − 2a − (n − 1)d] = (n/2)[2a + (3n − 1)d].
S₃ = (3n/2)[2a + (3n − 1)d] = 3(S₂ − S₁). Proved.
Teacher's Note:
1. Write all three sums with the formula.
2. Simplify S₂ − S₁ and compare with S₃.
Q28. If the sum of the first m terms of an AP is n and the sum of the first n terms is m (m ≠ n), prove that the sum of the first (m + n) terms is −(m + n). [4 marks]
View Answer & Explanation
Answer: From the two sums: 2a + (m − 1)d = 2n/m and 2a + (n − 1)d = 2m/n.
Subtracting: (m − n)d = 2(n² − m²)/(mn), so d = −2(m + n)/(mn).
2a + (m + n − 1)d = [2a + (m − 1)d] + nd = 2n/m − 2(m + n)/m = −2.
S₍ₘ₊ₙ₎ = ((m + n)/2)(−2) = −(m + n).
Teacher's Note:
1. Find d first by subtracting the two relations.
2. Then write 2a + (m + n − 1)d as the sum of two known parts.
Q29. If the sum of the first m terms of an AP equals the sum of the first n terms (m ≠ n), show that the sum of the first (m + n) terms is zero. [4 marks]
View Answer & Explanation
Answer: (m/2)[2a + (m − 1)d] = (n/2)[2a + (n − 1)d] gives 2a(m − n) + d(m² − n² − m + n) = 0.
Factorising: (m − n)[2a + (m + n − 1)d] = 0. Since m ≠ n, 2a + (m + n − 1)d = 0.
So S₍ₘ₊ₙ₎ = ((m + n)/2) × 0 = 0.
Teacher's Note:
1. Collect all terms involving a and d.
2. Cancel the non-zero factor (m − n).
Q30. If the mth term of an AP is 1/n and the nth term is 1/m, show that the sum of the first mn terms is (mn + 1)/2. [4 marks]
View Answer & Explanation
Answer: a + (m − 1)d = 1/n and a + (n − 1)d = 1/m. Subtracting: (m − n)d = (m − n)/(mn), so d = 1/(mn).
Then a = 1/n − (m − 1)/(mn) = 1/(mn).
S₍ₘₙ₎ = (mn/2)[2/(mn) + (mn − 1)/(mn)] = (mn + 1)/2.
Teacher's Note:
1. Find d first, then a.
2. Substitute into the sum formula.
Q31. A man saves ₹32 in the first year, ₹36 in the second year, and in this way increases his savings by ₹4 every year. In how many years will his total savings be ₹200? [4 marks]
View Answer & Explanation
Answer: a = 32, d = 4. (n/2)[64 + 4(n − 1)] = 200 gives n(30 + 2n) = 200, so n² + 15n − 100 = 0.
(n + 20)(n − 5) = 0, so n = 5 (the negative value is rejected).
It takes 5 years.
Teacher's Note:
1. Yearly savings form an AP.
2. Reject the negative root.
Q32. A man repays a loan of ₹3250 by paying ₹20 in the first month and increasing the payment by ₹15 every month. How long does he take to clear the loan? [4 marks]
View Answer & Explanation
Answer: a = 20, d = 15. (n/2)[40 + 15(n − 1)] = 3250 gives 15n² + 25n − 6500 = 0, i.e. 3n² + 5n − 1300 = 0.
D = 25 + 15600 = 15625 = 125², so n = (−5 + 125)/6 = 20.
He takes 20 months.
Teacher's Note:
1. Monthly payments form an AP.
2. Solve the quadratic and reject the negative root.
Q33. Find x if 1 + 6 + 11 + 16 + … + x = 148. [4 marks]
View Answer & Explanation
Answer: a = 1, d = 5. (n/2)[2 + 5(n − 1)] = 148 gives 5n² − 3n − 296 = 0.
D = 9 + 5920 = 5929 = 77², so n = (3 + 77)/10 = 8.
x = 1 + 7 × 5 = 36.
Teacher's Note:
1. First find the number of terms from the sum.
2. Then the last term is the nth term.
Q34. A sum of ₹700 is to be used to give seven cash prizes to students of a school, each prize being ₹20 less than the preceding prize. Find the value of each prize. [4 marks]
View Answer & Explanation
Answer: Let the first prize be a and d = −20. S₇ = (7/2)[2a − 120] = 700 gives 2a − 120 = 200, so a = 160.
The prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40.
Teacher's Note:
1. The prizes form a decreasing AP.
2. Check that the seven prizes add up to 700.
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