Show that a₁, a₂,…..a_n form an AP where is defined as below :

a_n = 9-5n

Also find the sum of the first 15 terms in each case.

Given a_n = 9-5n

So, a₁ = 9 – 5(1) = 4

a₂ = 9 -5(2) = -1

a₃ = 9 – 5(3) = -6

as of now a₁, a₂, a₃ have the same common difference

so given one is A.P

The sum of first 15 terms is given by (s_n) = [2a + (n-1) d]

= [ 2×4 +(15-1) (-5)]

= -465