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यदि A.P. $a _{1} a _{2}, a _{3}, \ldots$ के प्रथम 11 पदों का योगफल $0\left(a_{1} \neq 0\right)$ है और A.P., $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ का योगफल $ka _{1}$ है, तो $k$ बराबर है -
$\frac{121}{10}$
$-\frac{72}{5}$
$\frac{72}{5}$
$-\frac{121}{10}$
Solution
$a_{1}+a_{2}+a_{3}+\ldots \ldots+a_{11}=0$
$\Rightarrow\left(a_{1}+a_{11}\right) \times \frac{11}{2}=0$
$\Rightarrow \mathrm{a}_{1}+\mathrm{a}_{11}=0$
$\Rightarrow \mathrm{a}_{1}+\mathrm{a}_{1}+10 \mathrm{d}=0$
where d is common difference
$\Rightarrow \quad \mathrm{a}_{1}=-5 \mathrm{d}$
$a_{1}+a_{3}+a_{5}+\ldots \ldots+a_{23}$
$=\left(a_{1}+a_{23}\right) \times \frac{12}{2}=\left(a_{1}+a_{1}+22 d\right) \times 6$
$=\left(2 a_{1}+22\left(\frac{-a_{1}}{5}\right)\right) \times 6$
$=-\frac{72}{5} a_{1} \Rightarrow K=\frac{-72}{5}$
