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Find the sum of all the 11 terms of an AP whose middlemost term is 30.

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$\\$The middle term of an AP$ = \frac{n+1}{2} \\ $Here n $= 11 \\ $Hence a\textsubscript{6}$ = 30\\ a + 5d = 30 $ \ldots $ (1) (Using a\textsubscript{n} = a + (n - 1)d)\\ {{S}_{n}}=\frac{n}{2}\left[ 2a+\left( n-1 \right)d \right] put n=11\\ =\frac{11}{2}\left[ 2a+10d \right] \\ =\frac{11}{2}\left[ 2\left( a+5d \right) \right] \\ = 11 [30] (Using (1))\\ S\textsubscript{11} = 330\\$

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