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

Given:

The middle most term of an AP is 30.

To do:

We have to find the sum of all the 11 terms of the AP

Solution:

The number of terms $n=11$

This implies,

Middle most term $=\frac{n+1}{2}$th term

$=\frac{11+1}{2}$th term

$=6$th term

Let $a$ be the first term of the given AP and $d$ its common difference.

Therefore,

$a_6=30$

$a+5d=30$.........(i)

We know that,

Sum of $n$ terms of an AP $S_n=\frac{n}{2}[2a+(n-1)d]$

$S_{11}=\frac{11}{2}[2a+(11-1)d]$

$=\frac{11}{2}[2a+10d]$

$=\frac{11}{2}[2(a+5d)]$

$=11(30)$ [From (i)]

$=330$

The sum of all the 11 terms of the given AP is 330.

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