*Given an = 9-5n*

*to know the series, *

*a1 = 9-5(1) = 4*

*a2 = 9-5(2) = 9-10 = -1*

*a3 = 9-5(3) = 9-15 = -6*

*a4 = 9-5(4) = 9-20 = -11 ......*

*So the series is 4, -1, -6, -11 ... *

*common difference d = a2 - a1 = -1-4 = -5*

*So, the common difference d = -5 (we can check it by doing a3-a2, a4-a3 ... )*

*when we do with the a3-a2 and a4-a3 too, we get the common difference d=-5*

*So the {an} i.e, the series is in A.P*

*Then, by using*

*Sn = n/2 [2a +(n-1)d]*

*In above Calculation, We got that a = 4, d = -5, n=15 (as per problem) *

**he wasnt mentioned that in question, he mentioned that in comments*

*S15 = 15/2 [ 2(4)+(15-1)(-5)]*

*S15 = 15/2 [ 8+14(-5)]*

*S15 = 15/2 (-70+8)*

*S15 = 15/2 (-62)*

*S15 = 15(-31)*

*S15 = -465*

*therefore the sum of the first 15 numbers on the series that we find out is -465*