If the sum of the first 11 terms of an A.P.,a | vedclass

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(B) Given the sum of the first $11$ terms of the $A.P.$ is $0$: $S_{11} = \frac{11}{2}(2a_{1} + 10d) = 0$ $11(a_{1} + 5d) = 0 \Rightarrow a_{1} = -5d$

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$-\frac{72}{5}$

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(B) Given the sum of the first $11$ terms of the $A.P.$ is $0$: $S_{11} = \frac{11}{2}(2a_{1} + 10d) = 0$ $11(a_{1} + 5d) = 0 \Rightarrow a_{1} = -5d$ or $d = -\frac{a_{1}}{5}$. We need to find the sum of the $A.P.$ $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$. This is an $A.P.$ with $12$ terms,first term $A = a_{1}$ and common difference $D = 2d$. Sum $= \frac{12}{2}(2A + (12-1)D) = 6(2a_{1} + 11(2d)) = 6(2a_{1} + 22d)$. Substitute $d = -\frac{a_{1}}{5}$: Sum $= 6(2a_{1} + 22(-\frac{a_{1}}{5})) = 6(2a_{1} - \frac{22a_{1}}{5}) = 6(\frac{10a_{1} - 22a_{1}}{5}) = 6(-\frac{12a_{1}}{5}) = -\frac{72}{5}a_{1}$. Thus,$k = -\frac{72}{5}$.

If the sum of the first $11$ terms of an $A.P.$,$a_{1}, a_{2}, a_{3}, \ldots$ is $0$ $(a_{1} \neq 0)$,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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