Question

The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 40, find the common difference and the number of terms.

Answer 1

Given that, a = 5,

l = 45 and $S_n$ = 400

$S_n=\frac{n}{2}[a+l]$

$400=\frac{n}{2}(5+45)$

$400=\frac{n}{2}\left(50\right)$

n=16

$l=a+(n-1)d$

$45=5+(16-1)d$

$40=15d$

$d=\frac{40}{15}=\frac{8}{3}$