Question

for an AP a=11 and d=7. find the sum of the first 40 terms of the AP

Answer 1

Sn = 40/2 ( 2×11 + 39×7)

= 20 ( 22 + 273 )

= 20 × 295

= 5900

Answer 2

Given,

a=11

d=7

n=40

$S_{40}=\frac{n}{2}*[2a+(n-1)d]$

$\implies S_{40}=\frac{40}{2}*[22+39*7]$

$\implies S_{40}=20*295$

$\implies 5900$

The answer is 5900

Hope it helps.