Arithmetic Progressions Notes: Class 10 Maths Chapter 5

Class 10 Maths Chapter 5 Arithmetic Progressions covers the definition of an AP, the formula for the nth term, and the sum of the first n terms of an AP.

Overview of Topics & Subtopics (Click to Jump)

1. Arithmetic Progressions
2. nth Term of an AP
3. Sum of First n Terms of an AP

1. Arithmetic Progressions

Short Note:

• An Arithmetic Progression (AP) is a list of numbers in which each term (after the first) is obtained by adding a fixed number, called the common difference (d), to the preceding term.

• General form: a, a+d, a+2d, a+3d, ... where a is the first term.

2. nth Term of an AP

Short Note:

• The nth term of an AP is given by a_n = a + (n−1)d, where a is the first term and d is the common difference.

3. Sum of First n Terms of an AP

Short Note:

• Sum of the first n terms: S_n = n/2 · [2a + (n−1)d], or equivalently, S_n = n/2 · (a + a_n), where a_n is the last (nth) term.

NCERT Textbook Exercise Questions

Find the 10th term of the AP 2, 7, 12, 17, ...

Answer: Here a = 2, d = 5. a₁₀ = a + 9d = 2 + 9(5) = 2 + 45 = 47.

Which term of the AP 3, 8, 13, 18, ... is 78?

Answer: a = 3, d = 5. a_n = 78: 3 + (n−1)(5) = 78 ⇒ (n−1)(5) = 75 ⇒ n−1 = 15 ⇒ n = 16. So 78 is the 16th term.

Find the sum of the first 22 terms of the AP 8, 3, −2, ...

Answer: a = 8, d = −5, n = 22. S₂₂ = 22/2 × [2(8) + 21(−5)] = 11 × [16 − 105] = 11 × (−89) = −979.

How many terms of the AP 9, 17, 25, ... must be taken to give a sum of 636?

Answer: a = 9, d = 8. S_n = n/2 × [18 + (n−1)8] = 636 ⇒ n[18 + 8n − 8] = 1272 ⇒ n[8n+10] = 1272 ⇒ 8n²+10n−1272=0 ⇒ 4n²+5n−636=0. Solving this quadratic gives n = 12 (rejecting the negative root).

If the sum of the first n terms of an AP is given by S_n = 3n² − 4n, find the nth term.

Answer: a_n = S_n − S_n−1 = (3n²−4n) − [3(n−1)²−4(n−1)] = 6n − 7, after expanding and simplifying.