If the sum and product of the first three terms in an $AP$ are 33 and 1155 respectively, then the value of ${\mathbf{11}}^{\mathrm{th}}$ term in the Arithmetic progression is

Suppose that the three terms of an AP are $\left(a-d\right),a.\left(a+d\right)$ 

Given that the sum of the three terms is $33$

$\begin{array}{rcl}a-d+a+a+d& =& 33\\ 3a& =& 33\\ a& =& 11\end{array}$

Given that the product of three terms is $1155$

it implies that 

$\begin{array}{rcl}\left(a-d\right)\left(a\right)\left(a+d\right)& =& a\left(a^2-d^2\right)\\ & =& 11\left(121-d^2\right)\end{array}$

It implies 

 $\begin{array}{rcl}11\left(121-d^2\right)& =& 1155\\ 121-d^2& =& 105\\ d^2& =& 121-105\\ & =& 16\\ d& =& 4\end{array}$

Hence, the first term of the sequence is $a-d=11-4=7$

and the common difference is $d=\pm 4$

Therefore, 11th term of the sequence is 

$\begin{array}{rcl}{a}_{11}& =& a+10d\\ & =& 7+10\left(4\right)\\ & =& 7+40\\ & =& 47\end{array}$ or 

if the value of $d$ is $-4$, then $a-d=11+4=15$

Therefore, 11th term is 

     $\begin{array}{rcl}{a}_{11}& =& a+10d\\ & =& 15+10\left(-4\right)\\ & =& 15-40\\ & =& -25\end{array}$