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Explore structured practice materials through the CBSE Class 10 Mathematics Arithmetic Progressions Worksheet Set 16. Tailored for Class 10 learners, utilizing these Mathematics worksheets ensures thorough preparation and strengthens problem-solving accuracy before final school evaluations.
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Questions
Very Short Answer (1 mark)
Q1. The nth term of an AP is 7 − 4n. Find its common difference. [1 mark]
View Answer & Explanation
Answer: an+1 − an = [7 − 4(n + 1)] − (7 − 4n) = −4.
Teacher's Note:
1. The common difference is the difference of two consecutive terms.
2. The coefficient of n in a linear nth term is the common difference.
Q2. Which term of the AP 21, 18, 15, ... is zero? [1 mark]
View Answer & Explanation
Answer: a = 21, d = −3. 21 − 3(n − 1) = 0 gives n − 1 = 7, so n = 8. The 8th term is zero.
Teacher's Note:
1. Put an = 0 in the nth term formula.
2. The answer n must be a positive whole number.
Q3. For what value of p are 2p + 1, 13 and 5p − 3 three consecutive terms of an AP? [1 mark]
View Answer & Explanation
Answer: 2 × 13 = (2p + 1) + (5p − 3), so 26 = 7p − 2 and p = 4.
Teacher's Note:
1. Three consecutive terms satisfy 2b = a + c.
2. Putting p = 4 gives 9, 13, 17 with d = 4.
Q4. If \( a_n = \dfrac{n(n-3)}{n+4} \), find the 18th term of this sequence. [1 mark]
View Answer & Explanation
Answer: a18 = \( \dfrac{18 \times 15}{22} \) = \( \dfrac{135}{11} \).
Teacher's Note:
1. Substitute n = 18 in the given formula.
2. Cancel the common factor 2 to simplify the fraction.
Q5. If the sum of the first m terms of an AP is 2m² + 3m, what is its second term? [1 mark]
View Answer & Explanation
Answer: S1 = 5 and S2 = 8 + 6 = 14. So a2 = S2 − S1 = 9.
Teacher's Note:
1. The nth term equals Sn − Sn−1.
2. Here S1 is also the first term.
Q6. If the sum of the first p terms of an AP is ap² + bp, find its common difference. [1 mark]
View Answer & Explanation
Answer: a1 = S1 = a + b and S2 = 4a + 2b, so a2 = 3a + b. Then d = a2 − a1 = 2a.
Teacher's Note:
1. Find the first two terms from S1 and S2.
2. The common difference is their difference.
Short Answer I (2 marks)
Q7. Is −150 a term of the AP 17, 12, 7, 2, ...? [2 marks]
View Answer & Explanation
Answer: a = 17, d = −5. From 17 − 5(n − 1) = −150, n − 1 = 167/5, so n = 172/5. This is not a whole number, so −150 is not a term.
Teacher's Note:
1. Solve for n and test whether it is a positive integer.
2. A fractional n means the number is not in the AP.
Q8. In an AP, the sum of the first n terms is \( \dfrac{5n^2}{2} + \dfrac{3n}{2} \). Find its 20th term. [2 marks]
View Answer & Explanation
Answer: S20 = 1000 + 30 = 1030 and S19 = 902.5 + 28.5 = 931. So a20 = 1030 − 931 = 99.
Teacher's Note:
1. Use a20 = S20 − S19.
2. Calculate each sum carefully before subtracting.
Q9. Find the common difference of an AP whose first term is 4, last term is 49 and the sum of all terms is 265. [2 marks]
View Answer & Explanation
Answer: Sn = (n/2)(a + l) gives 265 = (n/2)(53), so n = 10. Then 49 = 4 + 9d and d = 5.
Teacher's Note:
1. First find n from the sum formula that uses the last term.
2. Then use the nth term formula to find d.
Q10. The sum of three numbers in an AP is 27 and their product is 405. Find the numbers. [2 marks]
View Answer & Explanation
Answer: Let the numbers be a − d, a, a + d. The sum 3a = 27 gives a = 9. The product 9(81 − d²) = 405 gives d² = 36, so d = ±6. The numbers are 3, 9, 15 (or 15, 9, 3).
Teacher's Note:
1. Taking a − d, a, a + d makes the sum depend on a only.
2. Both signs of d give the same three numbers in a different order.
Q11. Find k, if x = 1000 is the kth term of the AP 25, 50, 75, 100, ... [2 marks]
View Answer & Explanation
Answer: a = 25, d = 25. 1000 = 25 + (k − 1)25 gives k − 1 = 39, so k = 40.
Teacher's Note:
1. Put the given term equal to a + (k − 1)d.
2. Solve the linear equation for k.
Short Answer II (3 marks)
Q12. If m times the mth term of an AP is equal to n times its nth term, find the (m + n)th term. [3 marks]
View Answer & Explanation
Answer: m[a + (m − 1)d] = n[a + (n − 1)d]. So a(m − n) + d(m² − n² − m + n) = 0, that is (m − n)[a + (m + n − 1)d] = 0. As m ≠ n, a + (m + n − 1)d = 0, so the (m + n)th term is 0.
Teacher's Note:
1. Rearrange so that the factor (m − n) can be taken out.
2. The remaining factor is exactly the (m + n)th term.
Q13. If the 9th term of an AP is zero, prove that its 29th term is double its 19th term. [3 marks]
View Answer & Explanation
Answer: a + 8d = 0, so a = −8d. Then a19 = a + 18d = 10d and a29 = a + 28d = 20d. Hence a29 = 2 × a19.
Teacher's Note:
1. Express a in terms of d using the given condition.
2. Then compare the two required terms.
Q14. Which term of the progression 19, 18\( \tfrac{1}{5} \), 17\( \tfrac{2}{5} \), ... is the first negative term? [3 marks]
View Answer & Explanation
Answer: a = 19, d = −4/5. an = 19 − (4/5)(n − 1) < 0 gives n − 1 > 23.75, so n > 24.75. The first negative term is the 25th term (= −1/5).
Teacher's Note:
1. Form the inequality an < 0 and solve it.
2. Take the smallest whole number that satisfies it.
Q15. If the pth, qth and rth terms of an AP are x, y and z respectively, show that x(q − r) + y(r − p) + z(p − q) = 0. [3 marks]
View Answer & Explanation
Answer: Let the first term be a and the common difference d. Then x = a + (p − 1)d, y = a + (q − 1)d, z = a + (r − 1)d. The a terms give a[(q − r) + (r − p) + (p − q)] = 0. The d terms give d[(p − 1)(q − r) + (q − 1)(r − p) + (r − 1)(p − q)] . Hence the sum is 0.
Teacher's Note:
1. Write each of x, y and z in terms of a and d.
2. Expand carefully and notice that all terms cancel in pairs.
Q16. How many terms of the AP −6, −11/2, −5, ... are needed to give the sum −25? Explain the double answer. [3 marks]
View Answer & Explanation
Answer: a = −6, d = 1/2. Sn = (n/2)[−12 + (n − 1)/2] = −25 gives n² − 25n + 100 = 0, so n = 5 or n = 20. n = 5 or 20. The 15 terms from the 6th to the 20th run from −3.5 to 3.5 and cancel to give zero, so both 5 and 20 terms have sum −25.
Teacher's Note:
1. Form a quadratic in n and solve it.
2. Both roots are valid because the sum of the terms from the 6th to the 20th is zero.
Q17. If the pth term of an AP is \( \dfrac{1}{q} \) and the qth term is \( \dfrac{1}{p} \), show that the sum of pq terms is \( \dfrac{1}{2}(pq + 1) \). [3 marks]
View Answer & Explanation
Answer: a + (p − 1)d = 1/q and a + (q − 1)d = 1/p. Subtracting, (p − q)d = (p − q)/(pq), so d = 1/(pq) and then a = 1/(pq). Spq = (pq/2)[2/(pq) + (pq − 1)/(pq)] = (pq/2) × (pq + 1)/(pq) = \( \dfrac{1}{2} \)(pq + 1).
Teacher's Note:
1. Subtract the two equations to find d and then a.
2. Substitute both into the sum formula and simplify.
Long Answer (4 marks)
Q18. If the sum of the first 4 terms of an AP is 40 and that of the first 14 terms is 280, find the sum of its first n terms. [4 marks]
View Answer & Explanation
Answer: S4 gives 2a + 3d = 20 and S14 gives 2a + 13d = 40. Subtracting, 10d = 20, so d = 2 and a = 7. Sn = (n/2)[14 + 2(n − 1)] = n(n + 6).
Teacher's Note:
1. Form two linear equations in a and d.
2. Solve them and put the values in the sum formula.
Q19. The first and last terms of an AP are 8 and 350, and its common difference is 9. How many terms are there and what is their sum? [4 marks]
View Answer & Explanation
Answer: 350 = 8 + (n − 1)9 gives n − 1 = 38, so n = 39. S = (39/2)(8 + 350) = 39 × 179 = 6981. There are 39 terms.
Teacher's Note:
1. Find n from the last term first.
2. Then use the sum formula with the last term.
Q20. How many multiples of 4 lie between 10 and 250? Also find their sum. [4 marks]
View Answer & Explanation
Answer: The multiples are 12, 16, ..., 248. 248 = 12 + (n − 1)4 gives n = 60. The sum is (60/2)(12 + 248) = 30 × 260 = 7800. There are 60 multiples.
Teacher's Note:
1. List the first and last multiples inside the given range.
2. Then use the nth term and sum formulas.
Q21. Find the common difference of an AP whose first term is 5 and the sum of its first four terms is half the sum of the next four terms. [4 marks]
View Answer & Explanation
Answer: S4 = 2(10 + 3d) = 20 + 6d. The next four terms sum to S8 − S4 = 4(10 + 7d) − (20 + 6d) = 20 + 22d. The condition 20 + 6d = (1/2)(20 + 22d) gives 20 + 6d = 10 + 11d, so 5d = 10 and d = 2.
Teacher's Note:
1. Write the sum of the first four and of the next four in terms of d.
2. Solve the resulting linear equation.
Q22. Find the sum of the integers between 100 and 200 that are (i) divisible by 9, (ii) not divisible by 9. [4 marks]
View Answer & Explanation
Answer: (i) The multiples of 9 are 108, 117, ..., 198, so 198 = 108 + (n − 1)9 and n = 11. Their sum is (11/2)(108 + 198) = 1683. (ii) The sum of all integers 101 to 199 is (99/2)(101 + 199) = 14850. So the required sum is 14850 − 1683 = 13167.
Teacher's Note:
1. Handle the multiples of 9 as an AP and the full list of integers as another AP.
2. Subtract to get the numbers that are not divisible by 9.
Q23. Twenty-seven flags are fixed in a straight line 2 m apart, and a basket is placed at the middle flag. Ruchi starts from the basket, picks one flag at a time from there, and carries each flag to its place and returns to the basket. What total distance does she cover, and what is the longest distance she walks carrying a flag? [4 marks]
View Answer & Explanation
Answer: The middle flag is the 14th, so there are 13 flags on each side at distances 2, 4, ..., 26 m from the basket. For each flag she walks to the place and back. Total distance = 2 × 2 × (2 + 4 + ... + 26) = 4 × 182 = 728 m. The farthest flag is 26 m away, so the longest distance with a flag in hand is 26 m.
Teacher's Note:
1. The distances on each side form an AP with a = 2, d = 2 and 13 terms.
2. Remember to double for the two sides and again for the return walk.
Q24. Let S1, S2, S3 be the sums of n terms of three APs, each with first term 1 and common differences 1, 2 and 3 respectively. Prove that S1 + S3 = 2S2. [4 marks]
View Answer & Explanation
Answer: S1 = (n/2)[2 + (n − 1)] = n(n + 1)/2. S2 = (n/2)[2 + 2(n − 1)] = n². S3 = (n/2)[2 + 3(n − 1)] = n(3n − 1)/2. So S1 + S3 = [n² + n + 3n² − n]/2 = 2n² = 2S2.
Teacher's Note:
1. Write each sum with the sum formula.
2. Add the first and third and compare with twice the second.
Free study material for Mathematics
Free CBSE Practice Worksheets: Class 10 Mathematics Chapter 05 Arithmetic Progressions
Practice Exercises for Class 10 Mathematics Chapter 05 Arithmetic Progressions
Access structured practice worksheets for Chapter 05 Arithmetic Progressions aligned with the 2026 CBSE curriculum. These downloadable exercises for Class 10 Mathematics help students build accuracy and reinforce core concepts for upcoming school tests.
Step-by-Step Solutions and Practice Guidelines
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