Question

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

• A. 14
• B. 15
• C. 16
• D. 17

Answer 1

Answer: (B)

15

Answer 2

Given, AP is 2, 4, 6,... Here first term $a=2$ and common difference $d=4-2=2$ $\because$ ${{S}_{n}}=\frac{n}{2}[2a+(n-1)d]$
$\Rightarrow$ $240=\frac{n}{2}[2\times 2+(n-1)2]$
$\Rightarrow$ $480=4n+2n^2-2n]$
$\Rightarrow$ $480=2n+2{{n}^{2}}$
$\Rightarrow$ ${{n}^{2}}+n-240=0$
$\Rightarrow$ ${{n}^{2}}+16n-15n-240=0$
$\Rightarrow$ $(n-15)(n+16)=0$
$\Rightarrow$ $n=15,n=-16$ $n=-16$
is not possible.
$\therefore$ $n=15$