Using summation formula of G.P. having 1st term as a and common ratio as r we have,

$\frac{{a({r^3} – 1)}}{{(r – 1)}} = 42$ ……..(A)

Squaring the terms of the G.P. yields another G.P. having 1st term as $a^2$ and common ratio as $r^2 $. Thus,

$\frac{{{a^2}({r^6} – 1)}}{{({r^2} – 1)}} = 1092$

$ \Rightarrow \frac{{{a^2}({r^3} – 1)({r^3} + 1)}}{{(r – 1)(r + 1)}} = 1092$ Continue reading Find G.P. such that

$\sum\limits_{r = 1}^3 {{t_r}} = 42$

$\sum\limits_{r = 1}^3 {{{({t_r})}^2}} = 1092$ →