Let the first term $\alpha$ and the common ratio r of a geometric progression be positive integers. If the sum of squares of its first three terms is 33033, then the sum of these three terms is equal to

Given that the first term $a$ and common ratio $r$ of a geometric progression be positive integer. So, their 1st three terms are $a, a r, a r^2$ <br/><br/>According to the question, $a^2+a^2 r^2+a^2 r^4=33033$ <br/><br/>$$ \begin{aligned} \Rightarrow a^2\left(1+r^2+r^4\right) & =3 \times 7 \times 11 \times 11 \times 13 \\ & =3 \times 7 \times 13 \times 11^2 \end{aligned} $$ <br/><br/>$$ \begin{aligned} & \therefore \quad a^2=11^2 \\\\ & \Rightarrow \quad a=11 \\\\ & \text { and } 1+r^2+r^4=273 \\\\ & \Rightarrow r^2+r^4=272 \\\\ & \Rightarrow r^4+r^2-272=0 \\\\ & \Rightarrow\left(r^2+17\right)\left(r^2-16\right)=0 \\\\ & \Rightarrow r^2=-17(not ~possible),\\\\ & \Rightarrow r^2-16=0 \\\\ & \text { } \Rightarrow r= \pm 4 \\\\ & \Rightarrow r=4 ~~( \because r > 0) \end{aligned} $$ <br/><br/>So, sum of these first three terms is $a+a r+a r^2$ $=11+44+176=231$