What is the formula for the sum of n terms of an AP?

Sₙ = n/2 × [2a + (n − 1)d], or equivalently Sₙ = n/2 × (a + l) where l is the last term. Use the second form when you already know the first and last terms.

How do you find which term of an AP equals a given value?

Set aₙ = a + (n − 1)d equal to the given value and solve for n. If n comes out as a positive integer, that value is a term of the AP; otherwise it is not.

NCERT Solutions for Class 10 Maths Chapter 5 Arithmetic Progressions — Exercise Solutions

How to Use This Page

Complete NCERT Solutions for CBSE Class 10 Maths Chapter 5 — Arithmetic Progressions (AP): the nth-term formula, finding which term a value is, counting terms, and the sum of n terms.

Key Concepts (Quick Recap)

AP: a sequence with a constant common difference d = aₙ − aₙ₋₁.
nth term: aₙ = a + (n − 1)d.
Sum of n terms: Sₙ = n/2 [2a + (n − 1)d] = n/2 (a + l), where l = last term.

Exercise 5.1 — Identifying APs

Check whether each list forms an AP; if yes, write the common difference d.

2, 4, 8, 16, … → differences 2, 4, 8 are not equal → not an AP.
−10, −6, −2, 2, … → constant difference 4 → AP, d = 4.
3, 3 + √2, 3 + 2√2, … → constant difference √2 → AP, d = √2.
1, 3, 9, 27, … → ratios constant but differences 2, 6, 18 not equal → not an AP.

Exercise 5.2 — nth Term

Q1. Find the 10th term of the AP: 2, 7, 12, … a = 2, d = 5 → a₁₀ = 2 + 9(5) = 47.

Q2. Which term of the AP 3, 8, 13, 18, … is 78? a = 3, d = 5 → 3 + (n − 1)5 = 78 → 5(n − 1) = 75 → n − 1 = 15 → n = 16 (the 16th term).

Q3. Find the number of terms in the AP 7, 13, 19, …, 205. a = 7, d = 6, l = 205 → 205 = 7 + (n − 1)6 → 198 = 6(n − 1) → n − 1 = 33 → n = 34.

Q4. The 30th term of the AP 10, 7, 4, … is: a = 10, d = −3 → a₃₀ = 10 + 29(−3) = 10 − 87 = −77.

Q5. How many three-digit numbers are divisible by 7? They are 105, 112, …, 994 (a = 105, d = 7, l = 994). n = (994 − 105)/7 + 1 = 889/7 + 1 = 127 + 1 = 128.

Q6. Determine the AP whose 3rd term is 5 and 7th term is 9. a + 2d = 5 and a + 6d = 9 → subtract: 4d = 4 → d = 1, a = 3. AP: 3, 4, 5, 6, …

Exercise 5.3 — Sum of n Terms

Q1. Find the sum of the first 20 terms of the AP 2, 7, 12, … a = 2, d = 5 → S₂₀ = 20/2 [2(2) + 19(5)] = 10 [4 + 95] = 10 × 99 = 990.

Q2. Find the sum of the first 22 terms of the AP 8, 3, −2, … a = 8, d = −5 → S₂₂ = 22/2 [16 + 21(−5)] = 11 [16 − 105] = 11 × (−89) = −979.

Q3. Find the sum of the first 15 multiples of 8. 8, 16, …, 120 (a = 8, l = 120, n = 15) → S = 15/2 (8 + 120) = 15/2 × 128 = 15 × 64 = 960.

Q4. Find the sum of all two-digit numbers divisible by 3. 12, 15, …, 99 → n = (99 − 12)/3 + 1 = 30. S = 30/2 (12 + 99) = 15 × 111 = 1665.

Q5. Find the sum of the first 1000 positive integers. S = n(n + 1)/2 = 1000 × 1001 / 2 = 500500.

Q6. How many terms of the AP 9, 17, 25, … must be taken to give a sum of 636? a = 9, d = 8 → Sₙ = n/2 [18 + (n − 1)8] = 636 → n(4n + 5) = 636 → 4n² + 5n − 636 = 0 → (4n + 53)(n − 12) = 0 → n = 12.

Word Problems (Application)

Q. A sum of ₹700 is to be used to give seven cash prizes to students. If each prize is ₹20 less than its preceding prize, find the value of each prize. Let the largest prize be a; prizes form an AP with d = −20, n = 7, S₇ = 700. 700 = 7/2 [2a + 6(−20)] → 200 = 2a − 120 → 2a = 320 → a = 160. Prizes: ₹160, 140, 120, 100, 80, 60, 40.

Common Mistakes

Using a + nd instead of a + (n − 1)d for the nth term.
Forgetting d can be negative (decreasing AP).
In "how many terms," rejecting the valid positive-integer root of the quadratic and keeping the fraction.

Exam Tips

Two given terms → set up two equations (a + (m−1)d) to find a and d.
Use Sₙ = n/2 (a + l) when the last term is known — it is faster.
Always check that n is a positive integer; if not, the value is not a term/sum of the AP.

Frequently Asked Questions

The nth term is aₙ = a + (n − 1)d, where a is the first term and d is the common difference. The common difference d = a₂ − a₁ is the same between any two consecutive terms.