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

Given:

The first and the last terms of an A.P. are 5 and 45 respectively. The sum of all its terms is 400.

To do:

We have to find the common difference.

Solution:

Let the number of terms of the given A.P. be $n$, first term be $a$ and the common differnce be $d$.

First term $a=5$

Last term $l= 45$

Sum of all the terms $S_{n} =400$

We know that,

Sum of the $n$ terms$ S_{n} =\frac{n}{2}( a+l)$

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

$\Rightarrow 400=n(25)$

$\Rightarrow n=\frac{400}{25} =16$

Also,

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

Therefore,

On subtituting the values of $a$, $l$ and $n$, we get,

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

$\Rightarrow 15d=45-5=40$

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

Hence, the common difference of the given A.P. is $\frac{8}{3}$.