Question

Justify whether it is true to say that the sequence having following n^th term is an A.P.

a_n = 2n – 1

Answer 1

Given: a_n = 2n – 1.

First few terms:

a₁ = 2(1) – 1 = 1, 

a₂ = 2(2) – 1 = 3, 

a₃ = 5, ... so the sequence is 1, 3, 5, 7, ...

Compute difference between consecutive terms:

a_{n + 1} – a_n = [2(n + 1) – 1] – [2n – 1] 

= (2n + 1) – (2n – 1)

= 2

The difference a_{n + 1} – a_n is constant (equals 2) for all n, so the sequence has a fixed common difference.

Yes. The sequence with a_n = 2n – 1 is an arithmetic progression with first term a₁ = 1 and common difference d = 2.