Series

2+4=6+6=12+8=20+10=30....continuing in this manner we get 110+22=132

Let the $n^{th}$ term in the sequence be $a_n$ . Then we have:

$\begin{aligned} \{a_n \} & =2, 6, 12, 30... \\ & = 1 \times 2, 2 \times 3, 3 \times 4, 4 \times 5... n (n+1)... \\ \implies a_n & = n(n+1) \\ a_{1 1} & = 11\times 12 =\boxed {132}\end{aligned}$

The sequence is ( $a^{2}$ +a) or(axa+a) i.e.-->

1x1+1 , 2x2+2 , 3x3+3, ...

so the 11th term is 11x11+11=132 .