Question

The first and last term of an A.P are $5$ and $45$ respectively. If the sum of all the terms is $400,$ find the common difference.

Answer 1

Given: First term of an A.P, $a=5,n^{th}$ term, $a_n=45,$ Sum of ${}^{\prime }{n}^{\prime }$ terms, $S_n=400$

$n^{th}$ term of an A.P is given by, $a_n=a+(n-1)d$

$\Rightarrow 45=5+(n-1)d$

$\Rightarrow 45-5=(n-1)d$

$\Rightarrow (n-1)d=40\dots \dots \left(i\right)$

Sum of n terms of an A.P is given by $S_n=\frac{n}{2}(a+a_n)$

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

$\Rightarrow 800=50n$

$\Rightarrow n=\frac{800}{50}$

$\Rightarrow n=16$

Putting ${}^{\prime }{n}^{\prime }$ value in $eq.\left(i\right),$ we get

$(16-1)d=40$

$\Rightarrow d=\frac{40}{15}$

$\Rightarrow d=\frac{8}{3}$

Therefore, common difference $=\frac{8}{3}$