Question

The sum of the first four terms of an AP is 56 and the sum of the last four terms is 112. If its first term is 11, then find the number of terms.

Answer 1

Let there be n terms in the AP with first terms (a) =11 and common differenced.

Then, sum of first four terms=56

$\therefore 56=\frac{4}{2}[2a+(4-1\left)d\right][\because S_n\frac{n}{2}(2a+(n-1)d\left)\right]$

$\Rightarrow 2a+3d=28\Rightarrow 2\times 11+3d=28[\because a=11]$

$\Rightarrow 3d=6\Rightarrow d=2$

We have, sum of last four terms =11

$\Rightarrow 112=\frac{4}{2}(2a_n-3d)\Rightarrow 56=2(a+(n-1\left)d\right]-3d$ [1] [$\because$ sum of last n terms, $s_n=\frac{n}{2}(2a_n-(n-1\left)d\right]\left)\right]$

$\Rightarrow 56=2[11+(n-1\left)2\right]-3\times 2$ $[\because a={11}_{n}d=2]$

$\Rightarrow 56=22+4n-4-6\Rightarrow 44=4n\Rightarrow$n=11

Hence. there are 11 terms in the AP.