In a $G . P .$ of real numbers, it is given that $T_{1} + T_{2} + T_{3} + T_{4} = 30$
and $T_{1}^{2} + T_{2}^{2} + T_{3}^{2} + T_{4}^{2} = 340$ Determine the $G . P$ .

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Solution

$a (1 + r + r^{2} + r^{3}) = 30$ ..... $(1)$
$a^{2} (1 + r^{2} + r^{4} + r^{6}) = 340$
or $\frac{a (1 - r^{4})}{1 - r} = 30, \frac{a^{2} (1 - r^{8})}{1 - r^{2}} = 340$
Eliminate a by squaring $1 s t$ and dividing $2 n d$ .
$∴ \frac{a^{2} {(1 - r^{4})}^{2}}{(1 - {r)}^{2}} . \frac{1 - r^{2}}{a^{2} (1 - r^{8})} = \frac{900}{340} = \frac{45}{17}$
$\frac{1 - r^{4}}{1 + r^{4}} . \frac{1 + r}{1 - r} = \frac{45}{17}$
or $17 (1 + r + r^{2} + r^{3}) (1 + r) = 45 (1 + r^{4})$
or $17 (r^{4} + {2 r}^{3} + {2 r}^{2} + 2 r + 1) = 45 + 45 r^{4}$
or ${14 r}^{4} + {17 r}^{3} - 17 r^{2} - 17 r + 14 = 0$
Divide by $r^{2}$
$14 (r^{2} + \frac{1}{r^{2}}) - 17 (r + \frac{1}{r}) - 17 = 0$
Put $r + \frac{1}{r} = t$
$14 (t^{2} - 2) - 17 t - 17 = 0$ or $14 t^{2} - 17 t - 45 = 0$
$∴ t = \frac{17 \pm \sqrt{289 + 2520}}{28}$
or $r + \frac{1}{r} = \frac{17 \pm 53}{28} = \frac{5}{2}$ or $- \frac{9}{7}$
or $2 r^{2} - 5 r + 2 = 0$ and ${7 r}^{2} + 9 r + 7 = 0$
$∴ r = 2, \frac{1}{2}$ from $1 s t$ and $r$ is imaginary from $2 n d$
When $r = 2$ then from $(1), a = 2;$ when $r = \frac{1}{2}$ then $a = 16$ .
Hence the required $G . P s$ are $2, 4, 8, 16, 32, . .$ and $16, 8, 4, 2, 1,$

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