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

Given:

The first and the last terms of an AP are 7 and 49 respectively. If sum of all its terms is 420.

To do:

We have to find its common difference.

Solution:

Let $a$ be the first term, $d$ be the common difference and $n$ be the number of terms.

First term $a=7$

Last term $l=49$

Sum of the A.P. $S_{n}=420$

We know that,

$l=a_{n}=a+(n-1)d$

$\Rightarrow 49=7+(n-1)d$

$\Rightarrow ( n-1) d=49-7$

$\Rightarrow (n-1)d=42$........(i)

Sum of $n$ terms of an A.P. $S_{n} =\frac{n}{2}[ 2a+( n-1) d]$

$\Rightarrow 420=\frac{n}{2}[2(7)+42]$                  (From (i))

$\Rightarrow 420=n(7+21)$

$\Rightarrow n=\frac{420}{28}$

$n=15$

This implies,

$(15-1)d=42$

$\Rightarrow 14d=42$

$\Rightarrow d=\frac{42}{14}$

$\Rightarrow d=3$

Therefore, the common difference of the given A.P. is $3$.