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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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