Which term of A.P., 8, 11, 14, 17, .. is 272?

Formula Used.

d_n = a_{n + 1} – a_n

a_n = a + (n–1)d

In the above sequence,

a = 8;

d₁ = a₂–a₁ = 11–8 = 3

d₂ = a₃–a₂ = 14–11 = 3

d₃ = a₄–a₃ = 17–14 = 3

The difference in sequence is same and comes to be (3).

For any term of A.P to be 272

a_n = a + (n–1)d = 272

a_n = a + (n–1)d = 8 + (n–1)(3)

= 8 + 3n–3

= 5 + 3n

5 + 3n = 272

3n = 272–5 = 267

n = = 89

∴ The 89^th term of A.P has value 272