Question

The sum of fist four terms of an $A.P.$ is $56$ and sum of its last four terms is $112$. If its first term is $11$, then number of its terms is/are

• A. $10$
• B. $11$
• C. $12$
• D. $Noneofthese$

Answer 1

The correct option is B $11$

$S_4=56$ and $a=11$

Let $l$ be the last term then

$S_4=56$

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

$2\left[2a+3d\right]=56$

$2\times 11+3d=28$

$3d=28-22$

$3d=6$

$d=2$

$\frac{4}{2}\left[2l+\left(4-1\right)\left(-d\right)\right]=112$

$2\left[2l-3d\right]=112$

$2l-3\times 2=56$

$2l=56+6$

$2l=62$

$l=31$

$a+\left(n-1\right)d=31$

$11+\left(n-1\right)2=31$

$2\left(n-1\right)=20$

$n-1=10$

$n=11$

Sio, the number of terms is $11$