#### Answer

$24$

#### Work Step by Step

We can see that the sum of the first $n$ terms of an arithmetic sequence is $S_n=1092$ and $a_1=11$
There is a constant difference between the terms of: $d=3$
The sum of the first $n$ terms of an arithmetic sequence is given by:
$S_{n}= \dfrac{n}{2}\left[2a_{1}+(n-1) d\right] ..(1)$
Now, we plug in the above data into Equation-1 to find the number of terms, $n$.
$1092= \dfrac{n}{2}[(2)(11)+(3)(n-1)] \\ 2194=n [22+3n-3]\\ 2194=19n+3n^2 \\ 3n^2 +19n -2184=0$
Use the quadratic formula to find the roots of $n$.
$n=\dfrac{-19\pm\sqrt {(19)^2 -(4)(3) (-2184)}}{(2)(3)}=\dfrac{-19 \pm 163 }{6}$
Neglecting the negative terms, we have
$n=\dfrac{-19 +163 }{6}=24$
Therefore, we have $24$ terms.