Find the sum of all the 11 terms

Since, the total number of terms (n)=11                                                                               [odd]

$\therefore$ Middle most term $=\frac{(n+1)}{2}$ th term $=\left(\frac{11+1}{2}\right)$ th term $=6$ th term

Given that, $a_{6}=30$

$\Rightarrow \quad a+(6-1) d=30$

$\Rightarrow \quad a+5 d=30$ ...(1)

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

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

$=\frac{11}{2}(2 a+10 d)=11(a+5 d) \quad$ [from Eq. (i)]

$=11 \times 30=330$