Question

The nth term of an AP is given by (−4n + 15). Find the sum of first 20 terms of this AP?

Answer 1

Given, a_n = − 4n + 15

∴ a₁ = − 4 × 1 + 15 = − 4 + 15 = 11

a₂ = − 4 × 2 + 15 = − 8 + 15 = 7

a₃ = − 4 × 3 + 15 = − 12 + 15 = 3

a₄ = − 4 × 4 + 15 = − 16 + 15 = −1

It can be observed that

a₂ − a₁ = 7 − 11 = −4

a₃ − a₂ = 3 − 7 = −4

a₄ − a₃ = − 1 − 3 = −4

i.e., a_{k + 1} − a_k is same every time. Therefore, this is an A.P. with common difference as

−4 and first term as 11.

`S_n=n/2[2a+(n-1)d]`

`S_20=20/2[2(11)+(20-1)(-4)]`

`=10[22+19(-4)]`

`=10(20-76)`

`=10(-54)`

`=-540`