If the sum of first 11 terms of an A.P., a_{1 | vedclass

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Question

$a_{1}+a_{2}+a_{3}+\ldots \ldots+a_{11}=0$

$\Rightarrow\left(a_{1}+a_{11}ight) 	imes \frac{11}{2}=0$

$\Rightarrow \mathrm{a}_{1}+\mathrm{a}_{

Vedclass · Accepted Answer

$-\frac{72}{5}$

Vedclass · Answer

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

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 $A.P.$, $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ is $k a_{1},$ where $k$ is equal to

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