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 an $A.P.$ 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 an $A.P.$ is $0$: $S_{11} = \frac{11}{2} (2a_{1} + 10d) = 0$ $11(a_{1} + 5d) = 0 \Rightarrow a_{1} + 5d = 0 \Rightarrow d = -\frac{a_{1}}{5}$. We need to find the sum of the series $a_{1}, a_{3}, a_{5}, \ldots, a_{23}$. This is an $A.P.$ with $12$ terms,first term $A = a_{1}$ and common difference $D = 2d$. The sum $S'$ is given by: $S' = \frac{12}{2} [2A + (12-1)D] = 6 [2a_{1} + 11(2d)] = 6 [2a_{1} + 22d]$. Substituting $d = -\frac{a_{1}}{5}$: $S' = 6 [2a_{1} + 22(-\frac{a_{1}}{5})] = 6 [2a_{1} - \frac{22a_{1}}{5}] = 6 [\frac{10a_{1} - 22a_{1}}{5}] = 6 [-\frac{12a_{1}}{5}] = -\frac{72}{5} a_{1}$. 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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