Question

The first and the last terms of an AP are 5 and 45 respectively. If the sum of all its terms is 400, Find its common difference.

Answer 1

Let a be the first term and d be the common difference.
Given : a = 5
$T_n=45$
$S_n=400$
We know :
$T_n=a+(n-1)d$
$\Rightarrow 45=5+(n-1)d$
$\Rightarrow 40=(n-1)d$ ....(1)
And $S_n=\frac{n}{2}a+T_n$
$\Rightarrow 400=\frac{n}{2}(5+45)$
$\Rightarrow \frac{n}{2}=\frac{400}{50}$
$\Rightarrow n=2\times 8=16$
On substituting n = 16 in (1), we get
$40=(16-1)d$
$\Rightarrow 40=\left(15\right)d$
$\Rightarrow d=\frac{40}{15}=\frac{8}{3}$
Thus, the common difference is $\frac{8}{3}$