Question

Given that the sum of the first n term of an arithmetic progression is sn = n/2(18-8n). find the nth term

Answer 1

sum = n/2(18 - 8n)

=> n/2 {2a +(n - 1)d} = n/2 { 10 + (n -1) (-8)}

comparing LHS and RHS we get

2a = 10 and. d = -8

a = 5

therefore nth term = a + (n-1)d

= 5 + (n-1) (-8)

= 5 - 8n + 8

= 13 - 8n