Find the sum of all the 11 terms of an AP who | vedclass

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(B) The total number of terms is $n = 11$,which is odd.Therefore,the middle most term is the $\left(\frac{n+1}{2}\right)^{th}$ term,which is the $\lef

Vedclass · Accepted Answer

$330$

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(B) The total number of terms is $n = 11$,which is odd.Therefore,the middle most term is the $\left(\frac{n+1}{2}ight)^{th}$ term,which is the $\left(\frac{11+1}{2}ight)^{th} = 6^{th}$ term.Given that the $6^{th}$ term $a_6 = 30$.Using the formula for the $n^{th}$ term of an $AP$,$a_n = a + (n-1)d$,we have $a + 5d = 30$ ... $(i)$.The sum of $n$ terms of an $AP$ is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.For $n = 11$,$S_{11} = \frac{11}{2}[2a + (11-1)d] = \frac{11}{2}[2a + 10d]$.Factoring out $2$,we get $S_{11} = 11(a + 5d)$.Substituting the value from equation $(i)$,$S_{11} = 11 	imes 30 = 330$.

Find the sum of all the $11$ terms of an $AP$ whose middle most term is $30$.

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