A topic you 'somewhat know' is a topic you don't know yet. Be honest about your weak areas.

Home · Schooling · CBSE · Class 10 · Mathematics · Arithmetic Progressions · Exercise 5.3

NCERT Class 10 Mathematics Exercise 5.3 Solutions: Arithmetic Progressions

AI-written, checked on 23 Oct 2026

Solutions written by Shishya, checked — not the board's official answers

Step-by-step working for every question of Exercise 5.3 in NCERT's Class 10 Mathematics book, chapter “Arithmetic Progressions”, written by Shishya. Each question is in our own short words; open the NCERT chapter for the exact wording and figures.

NCERT chapter (PDF) ↗

How these were checked: the solutions were drafted with AI and then read and corrected on 23 Oct 2026. Every numerical final answer is recomputed from the question's own numbers by a calculator check that runs before each release; proofs and explanations were read, not machine-checked. The NCERT textbook is the authority — if something here disagrees with it, follow the book.

Tap “why?” under a step to see the textbook rule it uses — from Shishya's fixed list of NCERT Class 9-10 rules, not written by AI for this page. A step without one has no rule from the list yet.

Stuck on a step?

Ask the AI tutor about this chapter — it gives hints before answers — or snap a photo of your own question to get worked steps. You will be talking to an AI, not a person. For ages 13 and up; the first answer of each is free.

Ask the AI tutor about this chapter →📷 Snap a question →

Ex 5.3 Q1
Find the sums: (i) 2, 7, 12, … to 10 terms (ii) −37, −33, −29, … to 12 terms (iii) 0.6, 1.7, 2.8, … to 100 terms (iv) 1/15, 1/12, 1/10, … to 11 terms.
1. (i) The first term a is 2 and the common difference d is 7 − 2 = 5.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
2. (i) Using the sum formula S = n/2 [2a + (n − 1)d] with n = 10, we get S = 10/2 [4 + 9(5)].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
3. (i) Simplifying the expression gives S = 5 [4 + 45] = 5 × 49 = 245.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
4. (ii) The first term a is −37 and the common difference d is −33 − (−37) = 4.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
5. (ii) Using the sum formula with n = 12, we get S = 12/2 [2(−37) + 11(4)].
6. (ii) Simplifying the expression gives S = 6 [−74 + 44] = 6 × (−30) = −180.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. (iii) The first term a is 0.6 and the common difference d is 1.7 − 0.6 = 1.1.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
8. (iii) Using the sum formula with n = 100, we get S = 100/2 [2(0.6) + 99(1.1)].
9. (iii) Simplifying the expression gives S = 50 [1.2 + 108.9] = 50 × 110.1 = 5505.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
10. (iv) The first term a is 1/15 and the common difference d is 1/12 − 1/15 = 1/60.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
11. (iv) Using the sum formula with n = 11, we get S = 11/2 [2(1/15) + 10(1/60)].
12. (iv) Simplifying the expression gives S = 11/2 [2/15 + 1/6] = 11/2 [8/60 + 10/60] = 11/2 × 18/60 = 99/60 = 33/20.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
Answer: (i) 245; (ii) -180; (iii) 5505; (iv) 33/20
Report an error in this solution
Ex 5.3 Q2
Find the sums: (i) 7 + 10 1/2 + 14 + … + 84 (ii) 34 + 32 + 30 + … + 10 (iii) −5 + (−8) + (−11) + … + (−230).
1. (i) Here a = 7, d = 10.5 − 7 = 3.5, and l = 84.
2. (i) Using l = a + (n − 1)d, we get 84 = 7 + (n − 1)3.5, so 77 = (n − 1)3.5.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
3. (i) Solving for n gives n − 1 = 22, so n = 23.
4. (i) The sum is S = n/2 × (a + l) = 23/2 × (7 + 84) = 23/2 × 91 = 1046.5.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
5. (ii) Here a = 34, d = 32 − 34 = −2, and l = 10.
6. (ii) Using l = a + (n − 1)d, we get 10 = 34 + (n − 1)(−2), so −24 = −2(n − 1).
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
7. (ii) Solving for n gives n − 1 = 12, so n = 13.
8. (ii) The sum is S = n/2 × (a + l) = 13/2 × (34 + 10) = 13/2 × 44 = 13 × 22 = 286.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
9. (iii) Here a = −5, d = −8 − (−5) = −3, and l = −230.
10. (iii) Using l = a + (n − 1)d, we get −230 = −5 + (n − 1)(−3), so −225 = −3(n − 1).
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
11. (iii) Solving for n gives n − 1 = 75, so n = 76.
12. (iii) The sum is S = n/2 × (a + l) = 76/2 × (−5 + (−230)) = 38 × (−235) = −8930.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
Answer: (i) 1046.5; (ii) 286; (iii) -8930
Report an error in this solution
Ex 5.3 Q3
In an AP: (i) a = 5, d = 3, aₙ = 50: find n and Sₙ (ii) a = 7, a₁₃ = 35: find d and S₁₃ (iii) a₁₂ = 37, d = 3: find a and S₁₂ (iv) a₃ = 15, S₁₀ = 125: find d and a₁₀ (v) d = 5, S₉ = 75: find a and a₉ (vi) a = 2, d = 8, Sₙ = 90: find n and aₙ (vii) a = 8, aₙ = 62, Sₙ = 210: find n and d (viii) aₙ = 4, d = 2, Sₙ = −14: find n and a (ix) a = 3, n = 8, S = 192: find d (x) l = 28, S = 144, 9 terms: find a.
1. (i) Using aₙ = a + (n−1)d, 50 = 5 + (n−1)3, so 45 = 3(n−1) and n = 16.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
2. (i) Sₙ = n/2 [2a + (n−1)d] = 16/2 [10 + 45] = 8 × 55 = 440.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
3. (ii) Using a₁₃ = a + 12d, 35 = 7 + 12d, so 28 = 12d and d = 7/3.
4. (ii) S₁₃ = 13/2 (a + a₁₃) = 13/2 (7 + 35) = 13/2 × 42 = 273.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
5. (iii) Using a₁₂ = a + 11d, 37 = a + 33, so a = 4.
6. (iii) S₁₂ = 12/2 (a + a₁₂) = 6 (4 + 37) = 6 × 41 = 246.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
7. (iv) Using a₃ = a + 2d, 15 = a + 2d, so a = 15 − 2d.
8. (iv) S₁₀ = 10/2 [2a + 9d] = 125, so 5(2a + 9d) = 125 and 2a + 9d = 25.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
9. (iv) Substituting a = 15 − 2d gives 2(15 − 2d) + 9d = 25, so 30 + 5d = 25 and d = −1.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
10. (iv) a₁₀ = a + 9d = (15 − 2(−1)) + 9(−1) = 17 − 9 = 8.
11. (v) S₉ = 9/2 [2a + 8d] = 75, so 9(a + 4d) = 75 and a + 20 = 25/3.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
12. (v) a = 25/3 − 20 = −35/3.
13. (v) a₉ = a + 8d = −35/3 + 40 = 85/3.
14. (vi) Sₙ = n/2 [2a + (n−1)d] = 90, so n/2 [4 + 8(n−1)] = 90.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
15. (vi) Simplifying gives 4n² − 2n − 90 = 0, or 2n² − n − 45 = 0.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
16. (vi) Factoring gives (n−5)(2n+9) = 0, so n = 5 (since n > 0).
17. (vi) aₙ = a + (n−1)d = 2 + 4(8) = 34.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
18. (vii) Sₙ = n/2 (a + aₙ) = 210, so n/2 (8 + 62) = 210.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
19. (vii) 35n = 210, so n = 6.
20. (vii) d = (aₙ − a)/(n−1) = (62 − 8)/5 = 54/5.
21. (viii) Sₙ = n/2 (a + aₙ) = −14, so n/2 (a + 4) = −14.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
22. (viii) aₙ = a + (n−1)d, so 4 = a + 2(n−1), giving a = 6 − 2n.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
23. (viii) Substituting a gives n/2 (10 − 2n) = −14, so n(5 − n) = −14.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
24. (viii) n² − 5n − 14 = 0, so (n−7)(n+2) = 0, and n = 7.
25. (viii) a = 6 − 2(7) = −8.
26. (ix) Sₙ = n/2 [2a + (n−1)d] = 192, so 8/2 [6 + 7d] = 192.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
27. (ix) 4(6 + 7d) = 192, so 6 + 7d = 48 and 7d = 42.
28. (ix) d = 6.
29. (x) Sₙ = n/2 (a + l) = 144, so 9/2 (a + 28) = 144.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
30. (x) a + 28 = 32, so a = 4.
Answer: (i) n = 16, Sₙ = 440; (ii) d = 7/3, S₁₃ = 273; (iii) a = 4, S₁₂ = 246; (iv) d = -1, a₁₀ = 8; (v) a = -35/3, a₉ = 85/3; (vi) n = 5, aₙ = 34; (vii) n = 6, d = 54/5; (viii) n = 7, a = -8; (ix) d = 6; (x) a = 4
Report an error in this solution
Ex 5.3 Q4
How many terms of 9, 17, 25, … give a sum of 636?
1. The given arithmetic progression is 9, 17, 25, ...
2. The first term a is 9 and the common difference d is 17 − 9 = 8.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
3. The sum of n terms is given by the formula Sₙ = n/2 [2a + (n − 1)d].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
4. Substitute the known values into the formula: 636 = n/2 [2(9) + (n − 1)8].
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
5. Simplify the expression inside the brackets: 636 = n/2 [18 + 8n − 8].
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
6. Combine like terms: 636 = n/2 [8n + 10].
7. Multiply both sides by 2 to remove the fraction: 1272 = n(8n + 10).
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
8. Expand the right side: 1272 = 8n² + 10n.
  why?
  Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.
9. Rearrange into standard quadratic form: 8n² + 10n − 1272 = 0.
10. Divide the entire equation by 2 to simplify: 4n² + 5n − 636 = 0.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
11. Use the quadratic formula n = [−b ± √(b² − 4ac)] / 2a with a = 4, b = 5, c = −636.
  why?
  The quadratic formula x = (−b ± √(b² − 4ac)) / 2a comes from completing the square on ax² + bx + c = 0, so it gives both roots at once.
12. Calculate the discriminant: 5² − 4(4)(−636) = 25 + 10176 = 10201.
  why?
  For ax² + bx + c = 0 the discriminant D = b² − 4ac decides the roots: D > 0 two different real roots, D = 0 two equal real roots, D < 0 no real roots.
13. Find the square root of the discriminant: √10201 = 101.
  why?
  For ax² + bx + c = 0 the discriminant D = b² − 4ac decides the roots: D > 0 two different real roots, D = 0 two equal real roots, D < 0 no real roots.
14. Substitute back into the formula: n = [−5 ± 101] / 8.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
15. Take the positive value since the number of terms must be positive: n = (−5 + 101) / 8.
16. Calculate the final value: n = 96 / 8 = 12.
Answer: 12 terms
Report an error in this solution
Ex 5.3 Q5
An AP has first term 5, last term 45 and sum 400. Find the number of terms and d.
1. Use the sum formula S_n = n/2 (a + l) where a is the first term and l is the last term.
2. Substitute the given values: 400 = n/2 (5 + 45).
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
3. Simplify the equation: 400 = n/2 (50).
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
4. Calculate n: 400 = 25n, so n = 16.
5. Use the formula for the last term l = a + (n − 1)d.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
6. Substitute the known values: 45 = 5 + (16 − 1)d.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
7. Simplify: 45 = 5 + 15d.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
8. Solve for d: 40 = 15d, so d = 40/15 = 8/3.
Answer: 16 terms; d = 8/3
Report an error in this solution
Ex 5.3 Q6
An AP runs from 17 to 350 with d = 9. How many terms are there and what is their sum?
1. The first term a is 17 and the common difference d is 9.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
2. The last term l is given as 350.
3. Use the formula for the nth term: l = a + (n − 1)d.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
4. Substitute the values: 350 = 17 + (n − 1) × 9.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
5. Subtract 17 from both sides: 333 = (n − 1) × 9.
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
6. Divide by 9: n − 1 = 37.
7. Add 1 to find n: n = 38.
8. Use the sum formula: S = n/2 × (a + l).
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
9. Substitute the values: S = 38/2 × (17 + 350).
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
10. Simplify: S = 19 × 367.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
11. Calculate the product: S = 6973.
Answer: 38 terms, sum 6973
Report an error in this solution
Ex 5.3 Q7
Find the sum of the first 22 terms of an AP with d = 7 and 22nd term 149.
1. Given the common difference d = 7 and the 22nd term a₂₂ = 149.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
2. Use the formula for the nth term of an AP: aₙ = a + (n − 1)d.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
3. Substitute the known values into the formula: 149 = a + (22 − 1) × 7.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
4. Simplify the expression to find the first term: 149 = a + 21 × 7.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
5. Calculate the product: 149 = a + 147.
6. Solve for a: a = 149 − 147 = 2.
7. Use the formula for the sum of the first n terms: Sₙ = n/2 × [2a + (n − 1)d].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
8. Substitute n = 22, a = 2, and d = 7 into the sum formula.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
9. Sₙ = 22/2 × [2 × 2 + (22 − 1) × 7].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
10. Simplify the terms inside the bracket: Sₙ = 11 × [4 + 147].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
11. Add the values inside the bracket: Sₙ = 11 × 151.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
12. Calculate the final product: Sₙ = 1661.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
Answer: 1661
Report an error in this solution
Ex 5.3 Q8
Find the sum of the first 51 terms of an AP whose 2nd and 3rd terms are 14 and 18.
1. Given the 2nd term a₂ = 14 and the 3rd term a₃ = 18.
2. The common difference d is the difference between consecutive terms: d = a₃ − a₂ = 18 − 14 = 4.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
3. Using the formula for the nth term, a₂ = a + d, we find the first term: 14 = a + 4, so a = 10.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
4. We need the sum of the first 51 terms, so n = 51.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
5. Using the sum formula Sₙ = n/2 [2a + (n−1)d], substitute the values: S₅₁ = 51/2 [2(10) + (51−1)(4)].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
6. Simplify the expression inside the brackets: 20 + 50(4) = 20 + 200 = 220.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. Calculate the final sum: S₅₁ = 51/2 × 220 = 51 × 110 = 5610.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
Answer: 5610
Report an error in this solution
Ex 5.3 Q9
The first 7 terms of an AP add to 49 and the first 17 to 289. Find the sum of the first n terms.
1. The sum of the first 7 terms is given as 49, so S₇ = 49.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
2. Using the formula Sₙ = n/2 [2a + (n−1)d], we get 7/2 [2a + 6d] = 49.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
3. Simplifying this gives 2a + 6d = 14, which reduces to a + 3d = 7.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
4. The sum of the first 17 terms is given as 289, so S₁₇ = 289.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
5. Using the formula, we get 17/2 [2a + 16d] = 289.
6. Simplifying this gives 2a + 16d = 34, which reduces to a + 8d = 17.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. Subtracting the equation a + 3d = 7 from a + 8d = 17 gives 5d = 10.
8. Solving for d gives d = 2.
9. Substituting d = 2 into a + 3d = 7 gives a + 6 = 7, so a = 1.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
10. The sum of the first n terms is Sₙ = n/2 [2(1) + (n−1)2].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
11. Simplifying the expression gives Sₙ = n/2 [2 + 2n − 2] = n/2 [2n] = n².
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
Answer: Sₙ = n²
Report an error in this solution
Ex 5.3 Q10
Show that aₙ forms an AP and find the sum of the first 15 terms: (i) aₙ = 3 + 4n (ii) aₙ = 9 − 5n.
1. (i) Calculate the first term a₁ by substituting n = 1 into aₙ = 3 + 4n, giving a₁ = 3 + 4(1) = 7.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
2. (i) Calculate the second term a₂ by substituting n = 2, giving a₂ = 3 + 4(2) = 11.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
3. (i) Find the common difference d = a₂ − a₁ = 11 − 7 = 4.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
4. (i) Since the difference between consecutive terms is constant, the sequence forms an AP.
5. (i) Use the sum formula Sₙ = n/2 [2a + (n−1)d] with n = 15, a = 7, and d = 4.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
6. (i) Substitute values: S₁₅ = 15/2 [2(7) + (15−1)(4)] = 15/2 [14 + 56] = 15/2 [70].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
7. (i) Calculate the final sum: S₁₅ = 15 × 35 = 525.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
8. (ii) Calculate the first term a₁ by substituting n = 1 into aₙ = 9 − 5n, giving a₁ = 9 − 5(1) = 4.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
9. (ii) Calculate the second term a₂ by substituting n = 2, giving a₂ = 9 − 5(2) = −1.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
10. (ii) Find the common difference d = a₂ − a₁ = −1 − 4 = −5.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
11. (ii) Since the difference between consecutive terms is constant, the sequence forms an AP.
12. (ii) Use the sum formula Sₙ = n/2 [2a + (n−1)d] with n = 15, a = 4, and d = −5.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
13. (ii) Substitute values: S₁₅ = 15/2 [2(4) + (15−1)(−5)] = 15/2 [8 − 70] = 15/2 [−62].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
14. (ii) Calculate the final sum: S₁₅ = 15 × (−31) = −465.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
Answer: (i) d = 4, S₁₅ = 525; (ii) d = -5, S₁₅ = -465
Report an error in this solution
Ex 5.3 Q11
The sum of the first n terms of an AP is 4n − n². Find S₁, S₂, the 2nd term, the 3rd term, the 10th term and the nth term.
1. Given the sum of the first n terms is S_n = 4n − n².
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
2. To find S₁, substitute n = 1 into the formula: S₁ = 4(1) − 1² = 4 − 1 = 3.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
3. To find S₂, substitute n = 2 into the formula: S₂ = 4(2) − 2² = 8 − 4 = 4.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
4. The 2nd term a₂ is found by subtracting S₁ from S₂: a₂ = S₂ − S₁ = 4 − 3 = 1.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
5. To find the 3rd term a₃, first calculate S₃: S₃ = 4(3) − 3² = 12 − 9 = 3.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
6. Then, a₃ = S₃ − S₂ = 3 − 4 = −1.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
7. To find the 10th term a₁₀, first calculate S₁₀: S₁₀ = 4(10) − 10² = 40 − 100 = −60.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
8. Then, a₁₀ = S₁₀ − S₉. First find S₉ = 4(9) − 9² = 36 − 81 = −45.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
9. So, a₁₀ = −60 − (−45) = −60 + 45 = −15.
10. To find the nth term aₙ, use the formula aₙ = Sₙ − Sₙ₋₁.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
11. Substitute the expressions: aₙ = (4n − n²) − (4(n−1) − (n−1)²).
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
12. Expand the term for Sₙ₋₁: 4(n−1) − (n² − 2n + 1) = 4n − 4 − n² + 2n − 1 = −n² + 6n − 5.
  why?
  Expanding multiplies every term in one bracket by every term in the other: (a + b)(c + d) = ac + ad + bc + bd.
13. Subtract Sₙ₋₁ from Sₙ: (4n − n²) − (−n² + 6n − 5).
14. Simplify the expression: 4n − n² + n² − 6n + 5 = 5 − 2n.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
Answer: S₁ = 3; S₂ = 4; a₂ = 1; a₃ = -1; a₁₀ = -15; aₙ = 5 - 2n
Report an error in this solution
Ex 5.3 Q12
Find the sum of the first 40 positive integers divisible by 6.
1. The first positive integer divisible by 6 is 6, so the first term a = 6.
2. The common difference d is 6 because each subsequent term increases by 6.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
3. The number of terms n is given as 40.
4. Use the sum formula for an arithmetic progression: Sₙ = n/2 [2a + (n − 1)d].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
5. Substitute the values into the formula: S₄₀ = 40/2 [2(6) + (40 − 1)(6)].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
6. Simplify the expression inside the brackets: 2(6) = 12 and (39)(6) = 234.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. Add the terms inside the brackets: 12 + 234 = 246.
8. Multiply by n/2: 20 × 246 = 4920.
Answer: 4920
Report an error in this solution
Ex 5.3 Q13
Find the sum of the first 15 multiples of 8.
1. The first 15 multiples of 8 form an arithmetic progression: 8, 16, 24, ..., 120.
2. Here, the first term a = 8 and the number of terms n = 15.
3. The common difference d is 8.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
4. Using the sum formula S_n = n/2 [2a + (n-1)d], substitute the values.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
5. S_15 = 15/2 [2(8) + (15-1)(8)].
6. S_15 = 15/2 [16 + 14(8)].
7. S_15 = 15/2 [16 + 112].
8. S_15 = 15/2 [128].
9. S_15 = 15 × 64.
10. S_15 = 960.
Answer: 960
Report an error in this solution
Ex 5.3 Q14
Find the sum of the odd numbers between 0 and 50.
1. The odd numbers between 0 and 50 are 1, 3, 5, ..., 49.
2. This forms an arithmetic progression with first term a = 1 and common difference d = 2.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
3. The last term is 49, so we find n using 49 = 1 + (n − 1) × 2.
4. Solving gives 48 = 2(n − 1), so n − 1 = 24 and n = 25.
5. The sum is S = n/2 × (first + last) = 25/2 × (1 + 49).
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
6. S = 25/2 × 50 = 25 × 25 = 625.
Answer: 625
Report an error in this solution
Ex 5.3 Q15
A delay penalty is ₹200 on day 1, ₹250 on day 2, ₹300 on day 3 and so on. Find the penalty for a 30-day delay.
1. The daily penalties form an arithmetic progression with first term a = 200 and common difference d = 50.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
2. We need to find the penalty for the 30th day, which is the 30th term of the AP.
3. Using the formula for the nth term, aₙ = a + (n − 1)d.
  why?
  In an AP every term adds the same common difference d, so from the first term a there are (n − 1) steps of d to the nth term: aₙ = a + (n − 1)d.
4. Substitute the values: a₃₀ = 200 + (30 − 1) × 50.
  why?
  Putting the known values in place of the letters turns the formula or equation into plain arithmetic — the letters stood for exactly these values.
5. Calculate the term inside the bracket: 30 − 1 = 29.
6. Multiply by the common difference: 29 × 50 = 1450.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
7. Add the first term: 200 + 1450 = 1650.
8. The penalty for the 30th day is ₹1650.
9. To find the total penalty for a 30-day delay, we calculate the sum of the first 30 terms.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
10. Use the sum formula Sₙ = n/2 × [2a + (n − 1)d].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
11. Substitute the values: S₃₀ = 30/2 × [2 × 200 + (30 − 1) × 50].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
12. Simplify the bracket: 400 + 29 × 50 = 400 + 1450 = 1850.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
13. Multiply by n/2: 15 × 1850.
14. Calculate the final product: 15 × 1850 = 27750.
Answer: ₹27750
Report an error in this solution
Ex 5.3 Q16
₹700 is shared as seven prizes, each ₹20 less than the one before. Find each prize.
1. Let the seven prizes be a, a + d, a + 2d, a + 3d, a + 4d, a + 5d, and a + 6d.
2. Since each prize is ₹20 less than the one before, the common difference d is −20.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
3. The sum of the seven prizes is given as ₹700, so S₇ = 700.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
4. Using the sum formula Sₙ = n/2 [2a + (n − 1)d], we get 700 = 7/2 [2a + 6(−20)].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
5. Simplifying the equation gives 700 = 7/2 [2a − 120].
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
6. Dividing both sides by 7/2 yields 200 = 2a − 120.
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
7. Adding 120 to both sides gives 2a = 320.
  why?
  Doing the same thing to both sides of an equation (adding, subtracting, multiplying or dividing by the same non-zero number) keeps the two sides equal, so the equation stays true while the unknown is left on its own.
8. Dividing by 2 gives a = 160.
9. The prizes are 160, 160 − 20, 160 − 40, 160 − 60, 160 − 80, 160 − 100, and 160 − 120.
10. Thus, the prizes are ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, and ₹40.
Answer: ₹160, ₹140, ₹120, ₹100, ₹80, ₹60, ₹40
Report an error in this solution
Ex 5.3 Q17
Each of the 3 sections of Class I to XII plants as many trees as its class number. How many trees in all?
1. The number of trees planted by each class is equal to its class number.
2. So, Class I plants 1 tree, Class II plants 2 trees, and so on up to Class XII which plants 12 trees.
3. The total number of trees is the sum of the first 12 natural numbers: 1 + 2 + 3 + ... + 12.
4. We use the formula for the sum of the first n natural numbers, which is n(n + 1) / 2.
5. Here, n = 12, so the sum is 12(12 + 1) / 2.
6. This simplifies to 12 × 13 / 2.
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. Calculating this gives 156 / 2.
8. The result is 78 trees per section.
9. Since there are 3 sections, the total number of trees is 3 × 78.
10. Multiplying these gives 234 trees in all.
Answer: 234
Report an error in this solution
Ex 5.3 Q18
A spiral of 13 semicircles has radii 0.5 cm, 1.0 cm, 1.5 cm, … with centres alternately at A and B. Find its total length (π = 22/7).
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. The spiral consists of 13 semicircles with radii forming an arithmetic progression: 0.5, 1.0, 1.5, ..., 6.5 cm.
2. The length of a semicircle is given by the formula πr, where r is the radius.
3. The total length of the spiral is the sum of the lengths of all 13 semicircles: L = π(0.5 + 1.0 + 1.5 + ... + 6.5).
4. The sum of the radii is the sum of an AP with first term a = 0.5, last term l = 6.5, and number of terms n = 13.
5. Sum of radii = n/2 × (a + l) = 13/2 × (0.5 + 6.5) = 13/2 × 7 = 45.5 cm.
6. Total length L = π × 45.5 = (22/7) × 45.5.
7. L = (22 × 45.5) / 7 = 1001 / 7 = 143 cm.
Answer: 143 cm
Report an error in this solution
Ex 5.3 Q19
200 logs are stacked with 20 in the bottom row, 19 in the next, and so on. How many rows are there and how many logs are in the top row?
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. Let the number of rows be n.
2. The number of logs in the rows form an arithmetic progression: 20, 19, 18, ...
3. Here, first term a = 20 and common difference d = 19 − 20 = −1.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
4. The total number of logs is the sum of the AP, so S = 200.
5. Using the sum formula S = n/2 [2a + (n − 1)d], we get 200 = n/2 [2(20) + (n − 1)(−1)].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
6. Simplifying, 200 = n/2 [40 − n + 1], which gives 200 = n/2 (41 − n).
  why?
  Simplifying (combining like terms, cancelling common factors) changes how the expression looks, not its value, so the next step is easier.
7. Multiplying by 2, 400 = 41n − n².
8. Rearranging gives n² − 41n + 400 = 0.
9. Factoring the quadratic: (n − 16)(n − 25) = 0.
10. So n = 16 or n = 25.
11. If n = 25, the last term would be 20 + (25 − 1)(−1) = −4, which is impossible for a count of logs.
12. Therefore, n = 16 is the valid number of rows.
13. The number of logs in the top row is the 16th term: a₁₆ = 20 + (16 − 1)(−1).
14. Calculating, a₁₆ = 20 − 15 = 5.
Answer: 16 rows, 5 logs in the top row
Report an error in this solution
Ex 5.3 Q20
In a potato race the bucket is 5 m from the first of 10 potatoes placed 3 m apart in a line; each potato is fetched back to the bucket one by one. Find the total distance run.
Uses a figure printed in the NCERT book — open the chapter above to see it.
1. The bucket is 5 m from the first potato, so the distance to the 1st potato is 5 m.
2. The potatoes are placed 3 m apart, so the distance to the nth potato is 5 + (n − 1) × 3 m.
3. The player runs to the potato and back to the bucket, so the distance for the nth potato is 2 × [5 + (n − 1) × 3] m.
4. This forms an arithmetic progression with first term a = 10 m and common difference d = 6 m.
  why?
  In an arithmetic progression the difference between any term and the one before it is the same everywhere; that fixed gap is the common difference d = a₂ − a₁.
5. The total distance for 10 potatoes is the sum of 10 terms: S₁₀ = 10/2 × [2 × 10 + (10 − 1) × 6].
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
6. Simplifying the expression gives S₁₀ = 5 × [20 + 54] = 5 × 74.
  why?
  Sₙ = n/2 × (first term + last term) = n/2 [2a + (n − 1)d]: pairing the first and last terms, the second and second-last, and so on, gives pairs that all have the same total.
7. The total distance run is 370 m.
Answer: 370 m
Report an error in this solution

Other exercises in this chapter

Exercise 5.1 — 4 questions
Exercise 5.2 — 20 questions
Exercise 5.4 — 5 questions

← Arithmetic Progressions: notes, practice and all exercises

Saved for offline

Saves these solutions in this browser on this phone. No sign-in; nothing about you is saved or sent. A saved copy is not updated until you save it again. Parts that load from the server when you use them (the AI tutor, adaptive practice, sign-in) still need a connection.

NCERT Class 10 Mathematics Exercise 5.3 Solutions: Arithmetic Progressions

Ex 5.3 Q1

Ex 5.3 Q2

Ex 5.3 Q3

Ex 5.3 Q4

Ex 5.3 Q5

Ex 5.3 Q6

Ex 5.3 Q7

Ex 5.3 Q8

Ex 5.3 Q9

Ex 5.3 Q10

Ex 5.3 Q11

Ex 5.3 Q12

Ex 5.3 Q13

Ex 5.3 Q14

Ex 5.3 Q15

Ex 5.3 Q16

Ex 5.3 Q17

Ex 5.3 Q18

Ex 5.3 Q19

Ex 5.3 Q20