If the nth term of an A.P. is 9 — 5n, find the sum to first 15 terms.

Given: a_n = 9 – 5n

Taking n = 1,

a₁ = 9 – 5(1) = 9 – 5 = 4

Taking n = 2,

a₂ = 9 – 5(2) = 9 – 10 = -1

Taking n = 3,

a₃ = 9 – 5(3) = 9 – 15 = -6

Therefore the series is 4, -1, -6, …

So, a = 4, d = a₂ – a₁ = -1 – 4 = -5

Now, we have to find the sum of the first 15 terms of the AP

⇒ S₁₅ = 15 × (-31)

⇒ S₁₅ = -465

Hence, the sum of 15 terms is -465.