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Find G.P. such that

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

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

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$

Using (A), $42 \times \frac{{a({r^3} + 1)}}{{(r + 1)}} = 1092$

$ \Rightarrow a({r^2} - r + 1) = 26$

Given, $a (r^2 + r + 1) = 42$

$\therefore \frac{{{r^2} + r + 1}}{{{r^2} - r + 1}} = \frac{{42}}{{26}} = \frac{{21}}{{13}}$

$ \Rightarrow 8{r^2} - 34r + 8 = 0$ or $4{r^2} - 17r + 4 = 0$

$ \Rightarrow r = 4,1/4$

Using, $a = \frac{{26}}{{{r^2} - r + 1}};a = 2,32$

The 3 terms of G.P. are either 2, 8, 32 or 32, 8, 2

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