What is the nth term of an AP?

For an arithmetic progression with first term a and common difference d, the nth term is aₙ = a + (n − 1)d.

What is the sum of the first n terms of an AP?

Sₙ = n/2 [2a + (n − 1)d], or Sₙ = n/2 (a + l) when the last term l is known.

What is the relation between HCF and LCM of two numbers?

For two numbers a and b: HCF × LCM = a × b.

Tamil Nadu Class 10 Maths Chapter 2 - Numbers and Sequences

Q: What is Euclid's division lemma?

For positive integers a and b, there exist unique integers q and r such that a = bq + r, where 0 ≤ r < b. Repeatedly applying it gives the HCF.

Overview

This chapter blends number theory (Euclid's lemma, HCF, LCM) with sequences (arithmetic and geometric progressions). The formulae are fixed, so accuracy in substitution is what earns marks.

1. Euclid's Division Lemma and HCF

For positive integers a, b: a = bq + r, with 0 ≤ r < b.

To find HCF: divide, then replace the divisor with the remainder and repeat until the remainder is 0. The last non-zero remainder is the HCF.

HCF–LCM relation: HCF × LCM = a × b.

2. Arithmetic Progression (AP)

A sequence with a constant common difference d (d = a₂ − a₁).

nth term:  aₙ = a + (n − 1)d
Sum:       Sₙ = n/2 [2a + (n − 1)d]   or   Sₙ = n/2 (a + l)

3. Geometric Progression (GP)

A sequence with a constant common ratio r (r = a₂ / a₁).

nth term:  aₙ = a · rⁿ⁻¹
Sum:       Sₙ = a(rⁿ − 1)/(r − 1)   (r ≠ 1)

Worked Examples

Example 1 — HCF by Euclid's algorithm

Find the HCF of 867 and 255.

= 255 × 3 + 102
= 102 × 2 + 51
= 51 × 2 + 0
Last non-zero remainder = 51  →  HCF = 51

Example 2 — LCM using the relation

HCF of two numbers is 51 and their product is 221085. Find the LCM.

LCM = (a × b)/HCF = 221085 / 51 = 4335

Example 3 — nth term of an AP

Find the 18th term of 7, 13, 19, …

a = 7, d = 6 → a₁₈ = 7 + (18 − 1)(6) = 7 + 102 = 109

Example 4 — Sum of an AP

Find the sum of the first 25 terms of 5, 9, 13, …

a = 5, d = 4, n = 25
S₂₅ = 25/2 [2(5) + 24(4)] = 25/2 [10 + 96] = 25/2 (106) = 1325

Example 5 — Sum of a GP

Find the sum of 3 + 6 + 12 + … to 6 terms.

a = 3, r = 2, n = 6
S₆ = 3(2⁶ − 1)/(2 − 1) = 3(64 − 1) = 3 × 63 = 189

Example 6 — AP word problem

How many terms of the AP 24, 21, 18, … give a sum of 78?

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

Common Mistakes

Stopping Euclid's algorithm before the remainder reaches 0.
Using n instead of (n − 1) in the nth-term formula.
Forgetting the sign of a negative common difference.
Using the AP sum formula for a GP (and vice-versa).

Exam Tips & Revision Checklist

Memorise the four formulae (AP nth term & sum, GP nth term & sum).
Identify AP vs GP first (constant difference → AP; constant ratio → GP).
For "how many terms" problems, expect a quadratic in n — both roots may be valid.
Practised: Euclid HCF, HCF×LCM, AP nth/sum, GP nth/sum, word problems.

Frequently Asked Questions