If the sum of first 11 terms of an A.P., $a_{1} a_{2}, a_{3}, \ldots$ is $0\left(\mathrm{a}_{1} \neq 0\right)$, then the sum of the A.P., $\mathrm{a}_{1}, \mathrm{a}_{3}, \mathrm{a}_{5}, \ldots, \mathrm{a}_{23}$ is $\mathrm{ka}_{1}$, where $\mathrm{k}$ is equal to :
Correct Option: , 2
$a_{1}+a_{2}+a_{3}+\ldots+a_{11}=0$
$\Rightarrow\left(a_{1}+a_{11}\right) \times \frac{11}{2}=0$
$\Rightarrow a_{1}+a_{11}=0$
$\Rightarrow a_{1}+a_{1}+10 d=0$
where $d$ is common difference
$\Rightarrow a_{1}=-5 d$
$a_{1}+a_{3}+a_{5}+\ldots \ldots+a_{23}$
$=\left(a_{1}+a_{23}\right) \times \frac{12}{2}=\left(a_{1}+a_{1}+22 d\right) \times 6$
$=\left(2 \mathrm{a}_{1}+22\left(\frac{-\mathrm{a}_{1}}{5}\right)\right) \times 6$
$=-\frac{72}{5} a_{1} \Rightarrow K=\frac{-72}{5}$
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