Question

The first and the last terms of an A.P are $7$ and $49$ respectively. If the sum of all its terms is $420$, find its common difference.

Answer 1

Given:

$a=7,a_n=49,S_n=420$

For AP,

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

$420=\frac{n}{2}\times [7+49]$

$420=\frac{n}{2}\times \left[56\right]$

$n=15$.

As we know

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

$49=7+(15-1)d$

$49=7+(15-1)d$

$42=14d$

$d=3$.

So, the common difference is $3$.