Question

If the sum of first $11$ terms of an $A.P., a_{1} a_{2}, a_{3}, \ldots$ is $0\left(a_{1} \neq 0\right),$ then the sum of the $A.P., a _{1}, a _{3}, a _{5}, \ldots, a _{23}$ is $ka _{1},$ where $k$ is equal to :

• A. $\frac{121}{10}$
• B. $-\frac{72}{5}$
• C. $\frac{72}{5}$
• D. $-\frac{121}{10}$

Answer 1

Answer: (B)

$a_{1}+a_{2}+a_{3}+\ldots . .+a_{11}=0$
$\Rightarrow \left(a_{1}+a_{11}\right) \times \frac{11}{2}=0$
$\Rightarrow a_{1}+a_{11}=0$
$\Rightarrow a_{1}+a_{1}+10 d=0$
where $d$ is common difference
$\Rightarrow a_{1}=-5 d$
$a_{1}+a_{3}+a_{5}+\ldots \ldots+a_{23}$
$=\left(a_{1}+a_{23}\right) \times \frac{12}{2}=\left(a_{1}+a_{1}+22 d\right) \times 6$
$=\left(2 a_{1}+22\left(\frac{-a_{1}}{5}\right)\right) \times 6$
$=-\frac{72}{5} a _{1} \Rightarrow K =\frac{-72}{5}$