Question

If the sum of the first $2n$ terms of the $A.P.2,5,\mathrm{8..}$ is equal to the sum of the first $n$ terms of the $A.P.57,59,\mathrm{61...}$ then, $n=$

• A. $10$
• B. $12$
• C. $11$
• D. $13$

Answer 1

Answer: (C)

$11$

$S_{2n}$ $=n(4+(2n-1\left)3\right)$
$=n(6n+1)$ ...(i)
$S_n$ $=\frac{n}{2}\left(2\right(57)+(n-1\left)2\right)$
$=\frac{n}{2}(112+2n)$ ...(ii)
Now it is given that
$i=ii$
Hence
$n(6n+1)=\frac{n}{2}(112+2n)$
$2n(6n+1)=n(2n+112)$
$n(6n+1)=n(n+56)$
$n(6n+1)-n(n+56)=0$
$n[6n+1-n-56]=0$
$n(5n-55)=0$
$n=0$ or $n=\frac{55}{5}$
Hence $n=11$