Chapter 5: Arithmetic Progressions

Mathematics · CBSE Class 10 · NCERT Solutions

📖 Chapter 5 — Arithmetic Progressions (AP) (CBSE Class 10 Maths): Detailed notes — nth term and sum of an AP, with formulas and worked steps.

Chapter 5 — Arithmetic Progressions

🔢 1. What is an AP?

An Arithmetic Progression is a list of numbers in which each term is obtained by adding a fixed number d (the common difference) to the previous term. e.g. 2, 5, 8, 11, … (here a = 2, d = 3).

Common difference d = a₂ − a₁ = a₃ − a₂ = … (must be the same throughout).

📐 2. nth Term of an AP

aₙ = a + (n − 1)d

where a = first term, d = common difference, n = term number, aₙ = nth (general) term.

➕ 3. Sum of first n Terms

Sₙ = n/2 · [ 2a + (n − 1)d ] or Sₙ = n/2 · ( a + l )

where l = last term (aₙ). Also, aₙ = Sₙ − Sₙ₋₁.

🧠 Remember: d = a₂ − a₁ (same for all). aₙ = a + (n−1)d. Sₙ = n/2[2a+(n−1)d] = n/2(a+l). Number of terms n = (l − a)/d + 1.

⭐ Exam Tips: ① Check if a list is an AP (constant d). ② Find aₙ, or n, or a, or d using aₙ = a+(n−1)d. ③ Find Sₙ. ④ "Which term is …?" → solve aₙ = value for n. ⑤ Word problems (savings, salary, seats).

📘 EXERCISE 5.1

1In which situations does the list of numbers make an AP, and why?

Check whether the difference between consecutive terms is the same. (i) Taxi fare: 15, 23, 31, 39, … → d = 8 each time (constant) → AP. (ii) Air in cylinder: each time 3/4 of the previous remains → terms are in a ratio, not a fixed difference → Not an AP. (iii) Digging a well: 150, 200, 250, … → d = 50 (constant) → AP. (iv) Money at compound interest: multiplied by 1.08 each year → a ratio → Not an AP.

📗 EXERCISE 5.2 (nth Term)

1Fill in the blanks (a = first term, d = common difference, aₙ = nth term).

Use aₙ = a + (n−1)d for each row. (i) a=7, d=3, n=8 → a₈ = 7 + 7×3 = 28. (ii) a=−18, n=10, a₁₀=0 → 0 = −18 + 9d → 9d = 18 → d = 2. (iii) d=−3, n=18, a₁₈=−5 → −5 = a + 17(−3) → a = −5 + 51 → a = 46. (iv) a=−18.9, d=2.5, aₙ=3.6 → 3.6 = −18.9 + (n−1)2.5 → 22.5 = (n−1)2.5 → n−1 = 9 → n = 10. (v) a=3.5, d=0, n=105 → a₁₀₅ = 3.5 + 104×0 = 3.5.

2The 30th term of the AP 10, 7, 4, … is (A) 97 (B) 77 (C) −77 (D) −87.

Given: a = 10, d = 7 − 10 = −3, n = 30 Formula: aₙ = a + (n−1)d. a₃₀ = 10 + (30−1)(−3) = 10 + 29(−3). = 10 − 87 = −77. ∴ Correct option: (C) −77

📙 EXERCISE 5.3 (Sum of n Terms)

1Find the sum of the AP 2, 7, 12, … to 10 terms.

Given: a = 2, d = 7 − 2 = 5, n = 10 Formula: Sₙ = n/2 · [2a + (n−1)d]. S₁₀ = 10/2 · [2(2) + (10−1)(5)] = 5 · [4 + 45]. = 5 × 49. ∴ S₁₀ = 245

3In an AP, given a = 5, d = 3, aₙ = 50; find n and Sₙ.

Given: a = 5, d = 3, aₙ = 50 Find n using aₙ = a + (n−1)d → 50 = 5 + (n−1)3. 45 = (n−1)3 → n−1 = 15 → n = 16. Find Sₙ using Sₙ = n/2 · (a + aₙ) = 16/2 · (5 + 50). = 8 × 55. ∴ n = 16 and S₁₆ = 440

📕 EXERCISE 5.4 (Optional)

1Which term of the AP 121, 117, 113, … is its first negative term?

Given: a = 121, d = 117 − 121 = −4 We need the first term that is negative, i.e. aₙ < 0. a + (n−1)d < 0 → 121 + (n−1)(−4) < 0. 121 − 4n + 4 < 0 → 125 < 4n → n > 31.25. The smallest whole number greater than 31.25 is 32. ∴ The 32nd term is the first negative term.

Extra practice (4× the textbook) — full step-by-step for exams.

🎯 Extra MCQ

1The 7th term of AP 3, 8, 13, … is —

a=3, d=5, n=7a₇ = 3 + (7−1)5 = 3 + 30.∴ a₇ = 33

2Which term of AP 5, 8, 11, … is 320?

a=5, d=3, aₙ=320320 = 5 + (n−1)3 → 315 = 3(n−1) → n−1 = 105.∴ n = 106 (the 106th term)

3Is 301 a term of the AP 5, 11, 17, …?

a=5, d=6301 = 5 + (n−1)6 → 296 = 6(n−1) → n−1 = 49.33…, not a whole number.∴ 301 is NOT a term of this AP

📝 Short Answer

4Find the sum of first 20 terms of AP 1, 4, 7, 10, …

a=1, d=3, n=20S₂₀ = 20/2 · [2(1) + (20−1)3] = 10 · [2 + 57].= 10 × 59.∴ S₂₀ = 590

5Find the 11th term from the last term of the AP 10, 7, 4, …, −62.

last term l=−62, d=−3nth term from end = l − (n−1)d = −62 − (11−1)(−3).= −62 + 30.∴ 11th term from the last = −32

6How many two-digit numbers are divisible by 3?

AP: 12, 15, …, 99; a=12, d=3, l=99n = (l−a)/d + 1 = (99−12)/3 + 1 = 29 + 1.∴ 30 two-digit numbers are divisible by 3

7Find d if the 3rd term is 5 and the 7th term is 9.

a+2d=5 and a+6d=9 → subtract: 4d=4.∴ d = 1 (and a = 3)

📖 Long Answer

8The sum of the first 14 terms of an AP is 1050 and its first term is 10. Find the 20th term.

S₁₄=1050, a=10, n=14S₁₄ = 14/2[2(10)+(14−1)d] = 1050 → 7[20+13d] = 1050.20 + 13d = 150 → 13d = 130 → d = 10.a₂₀ = a + 19d = 10 + 19(10).∴ a₂₀ = 200

9How many terms of the AP 24, 21, 18, … must be taken so that their sum is 78?

a=24, d=−3, Sₙ=78n/2[2(24)+(n−1)(−3)] = 78 → n/2[48−3n+3] = 78.n(51−3n) = 156 → 3n²−51n+156 = 0 → n²−17n+52 = 0.(n−4)(n−13)=0 → n = 4 or n = 13.∴ n = 4 or 13 (both give sum 78)

10Find the sum: 34 + 32 + 30 + … + 10.

a=34, d=−2, l=10n = (l−a)/d + 1 = (10−34)/(−2) + 1 = 12 + 1 = 13.Sₙ = n/2(a+l) = 13/2(34+10) = 13/2 × 44.∴ Sum = 286

11The 17th term of an AP exceeds its 10th term by 7. Find the common difference.

a₁₇ − a₁₀ = 7 → [a+16d] − [a+9d] = 7 → 7d = 7.∴ d = 1

12A sum of ₹700 is to be given as 7 cash prizes, each ₹20 less than the one before. Find each prize.

n=7, Sₙ=700, d=−20700 = 7/2[2a + 6(−20)] → 200 = 2a − 120 → 2a = 320 → a = 160.Prizes: 160, 140, 120, 100, 80, 60, 40.∴ ₹160, 140, 120, 100, 80, 60, 40