If the sum of first four terms of an A.P. is 40 and that of first 14 terms is 280. Find the sum of its first n terms.

Given:

The sum of first four terms of an A.P. is 40 and that of first 14 terms is 280.

To do:

We have to find the sum of its first $n$ terms.

Solution:

Let the first term be $a$ and the common differnce be $d$.

We know that,

Sum of $n$ terms$ S_{n} =\frac{n}{2}(2a+(n-1)d)$

$S_{4}=\frac{4}{2}[2(a)+(4-1)d]$

$40=2(2a+3d)$

$20=2a+3d$

$2a=20-3d$......(i)

$S_{14}=\frac{14}{2}[2(a)+(14-1)d]$

$280=7(2a+13d)$

$40=2a+13d$

$20-3d+13d=40$       (From (i))

$10d=40-20$

$d=\frac{20}{10}$

$d=2$

This implies,

$2a=20-3(2)$

$2a=20-6$

$a=\frac{14}{2}$

$a=7$

The sum of $n$ terms $S_n=\frac{n}{2}(2a+(n-1)d)$

$S_n=\frac{n}{2}[2(7)+(n-1)2]$

$=n(7+n-1)$

$=n(n+6)$

$=n^2+6n$

Hence, the sum of $n$ terms is $n^2+6n$.