Question

If the sum of the first $2n$ terms of the A.P $2,5,8,...$ is equal to the sum of the first $n$ terms of the A.P $57,59,61,...$ then $n$ equals

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

Answer 1

The correct option is C $11$

In the given A.P sequence $2,5,\mathrm{8...}$ the first term is $a=2$ and the common difference is $d=3$.

Also, $S_{2n}=\frac{2n}{2}[2a+(2n-1\left)d\right]$ ...$\left(1\right)$

$\Rightarrow S_{2n}=\frac{2n}{2}\left[2\right(2)+(2n-1\left)3\right]$

$\Rightarrow S_{2n}=\frac{2n}{2}[4+(2n-1\left)3\right]$

Also,the first term and the common difference of the sequence $57,59,61,...$ are $a=57$ and $d=2$ respectively.

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

$\Rightarrow S_n=\frac{n}{2}\left[2\right(57)+(n-1\left)2\right]$

$=\frac{n}{2}[114+(n-1\left)2\right]$ ....$\left(2\right)$

We have given that both the sequence are equal, therefore from equation

$\left(1\right)$ and $\left(2\right)$, we get

$2\frac{n}{2}[4+(2n-1\left)3\right]=\frac{n}{2}[114+(n-1\left)2\right]$

$8+(2n-1)6=114+2n-2]$

$\Rightarrow 8+12n-6=2n+112$

$\Rightarrow 12n-2n=112-2$

$\Rightarrow 10n=110$

$\Rightarrow n=11$