Question

What is the sum to n terms of the following series: 2,4,6,8,…, ( $n^{th}$ term)?

• A. $n(n+1)$
• B. $n(n-1)$
• C. $n\times n$
• D. $(n+1)(n+2)$

Answer 1

The correct option is A

$n(n+1)$

The first term, a = 2

The common difference, d = $t_2$ – $t_1$ = 4 – 2 = 2.

Sum to ‘n’ terms,

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

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

$\Rightarrow S_n$ = $\frac{n}{2}(4+2n-2)$

$\Rightarrow S_n$ = $\frac{n}{2}(2n+2)$

$\Rightarrow S_n$ = $n(n+1)$