Question

When the sum of the first ten terms of an $A.P.$ is four times the sum of the first five terms. Then the $k^{th}$ terms is

• A. $a(2k+1)$
• B. $a(2k-1)$
• C. $2k+1$
• D. $2k+3$

Answer 1

The correct option is B $a(2k-1)$

Let ${}^{\prime }{a}^{\prime }$ be the first term and ${}^{\prime }{d}^{\prime }$ be the common difference of an $A.P$
Using sum formula $=$ $\frac{n}{2}(2a+(n-1\left)d\right)$
Given sum of first ten terms $=4\times$ sum of first five terms
$\Rightarrow \frac{10}{2}[2a+(10-1\left)d\right]=4\times \frac{5}{2}[2a+(5-1\left)d\right]$
$\Rightarrow 2a+9d=4a+8d$
$\Rightarrow 2a=d$ ........$\left(1\right)$
$k^{th}$ term of an $A.P.$ is $a+(K-1)d$
$\Rightarrow a+(k-1)2a$ ....By $\left(1\right)$
$\Rightarrow 2ak-a=a(2k-1)$