Question

If the sum to $2n$ terms of an AP $2,5,8,11,...$is equal to the sum of $n$ terms of an AP $57,59,61,63,...,$then $n$ is equal to

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

Answer 1

The correct option is B

$11$

Explanation for the correct option.

Step 1. Find the sum of $2n$ terms for the first AP.

For the AP $2,5,8,11..$ the first-term is $a=2$ and the common difference $d$ is $5-8=3$. So the sum of $2n$ terms is given as:

$\begin{array}{rcl}{S}_{2n}& =& \frac{2n}{2}\left[2a+\left(2n-1\right)d\right]\\ & =& n\left[2\times 2+\left(2n-1\right)\times 3\right]\left[a=2,d=3\right]\\ & =& n\left[4+6n-3\right]\\ & =& n\left(6n+1\right)\end{array}$

Step 2. Find the sum of $n$ terms for the second AP.

For the AP$57,59,61,63,.$ the first-term is $a=57$ and the common difference $d$ is $59-57=2$. So the sum of $n$ terms is given as:

$\begin{array}{rcl}{S}_n& =& \frac{n}{2}\left[2a+\left(n-1\right)d\right]\\ & =& \frac{n}{2}\left[2\times 57+\left(n-1\right)\times 2\right]\left[a=57,d=2\right]\\ & =& \frac{n}{2}\left[114+2n-2\right]\\ & =& \frac{n}{2}\times 2\left(n+56\right)\\ & =& n\left(n+56\right)\end{array}$

Step 3. Form the equation and solve for $n$.

It is given that the sum of $2n$ terms for the first AP is equal to the sum of $n$ terms of the second AP. So

$$\begin{gathered} n\left(6n+1\right)=n\left(n+56\right) \\ \Rightarrow 6n+1=n+56 \\ \Rightarrow 6n-n=56-1 \\ \Rightarrow 5n=55 \\ \Rightarrow n=11 \end{gathered}$$

Hence the correct option is B.