Question

The nth term of a progression is (3n + 5). Prove that this progression is an arithmetic progression. Also, find its 6th term.

Answer 1

Let T_n = 3n + 5

Then `T_((n − 1)) = 3(n − 1) + 5`

= 3n + 2

So the difference between consecutive terms is T_n − T(n-1)

= (3n + 5) − (3n + 2)

= 3

Hence, the progression is an arithmetic progression with common difference d = 3.

T₆ = 3 × 6 + 5

= 18 + 5

= 23