Question

Question 29
Find the sum of all the 11 terms of an AP whose middle most term is 30.

Answer 1

Total number of terms (n) = 11 [odd]
$\therefore Middlemostterm=\frac{(n+1)}{2}thterm=\left(\frac{11+1}{2}\right)thterm=6thterm$

Hence, $a_6=30$
$\Rightarrow a+(6-1)d=30[\because a_n=a+(n-1\left)d\right]$
$\Rightarrow a+5d=\mathrm{30...}\left(i\right)$

Sum of terms of an AP,
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$\therefore S_{11}=\frac{11}{2}[2a+(n-1\left)d\right]$
$=\frac{11}{2}(2a+10d)=11(a+5d)$
$=11\times 30=330\left[fromeq\right(i\left)\right]$