Question

If the sum to $2n$ terms of the $A.P. \text{ }2,\text{ }5,\text{ }8,11,...$ is equal to the sum to $n$ terms of the $\text{ }57,\text{ }59,\text{ }61,\text{ }63,\text{ }...\text{ },$ then $n$ =

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

Answer 1

Answer: (B)

$11$

Answer 2

Let sum of 2n terms of the $AP\,\,\,57,59,61,63$
is ${{S}_{n}}$ .
$\therefore$ ${{S}_{n}}=\frac{n}{2}[2\times 57+(n-1)2]$
$\frac{n}{2}(2n+112)$
According to question ${{S}_{2n}}={{S}_{n}}$
$\Rightarrow$ $n(6n+1)=\frac{n}{2}(2n+112)$
$\Rightarrow$ $12n+2=2n+112$
$\Rightarrow$ $10n=110$
$\Rightarrow$ $n=11$