Question

If the sum to first $n$ terms of an AP $2,4,6,...$ is $240$, then the value of $n$ is

• A. $14$
• B. $15$
• C. $16$
• D. $17$

Answer 1

The correct option is B

$15$

Explanation for the correct option.

Step 1. Find the sum of $n$ terms.

For the AP $2,4,6,...$ the first term is $a=2$ and the common difference $d$ is $4-2=2$. So the sum of $n$ terms is given as:

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

Step 2. Form the equation and solve for $n$.

It is given that the sum of $n$ terms is $240$, so

$$\begin{gathered} S_n=240 \\ \Rightarrow n^2+n=240 \\ \Rightarrow n^2+16n-15n-240=0 \\ \Rightarrow n\left(n+16\right)-15\left(n-16\right)=0 \\ \Rightarrow \left(n+16\right)\left(n-15\right)=0 \end{gathered}$$

So either $n=-16$ or $n=15$.

But the number of terms cannot be negative, so $n=-16$ is rejected.

Hence, the correct option is B.