Answer
$24$
Work Step by Step
The sum $S_{n}$ of the first $n$ terms of $\left\{a_{n}\right\}$:
$S_{n}=\displaystyle \frac{n}{2}\left[2a_{1}+(n-1)d\right]$
$ 1092=\displaystyle \frac{n}{2}\left[2(11)+3(n-1)\right]\quad$ ... solve for n
$2194=n[22+3n-3]$
$2194=19n+3n^{2}$
$3n^{2}+19n-2184=0$
... use the quadratic formula
$n=\displaystyle \frac{-19\pm\sqrt{19^{2}-4(3)(-2184)}}{2(3)}$
... discard the negative n....
$n=\displaystyle \frac{-19+163}{6}=24$
