Determine two $G . P . s$ which satisfy the following conditions:
$T_{1} + T_{2} + T_{3} + T_{4} = 30$ and $T_{5} + T_{6} + T_{7} + T_{8} = 480$ Also prove the following for both:
(i) $S_{6} - S_{4} = 96$
(ii) $\frac{T_{n}}{T_{n}^{'}} = \frac{1}{3} {(- 1)}^{n}$

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Solution

$S_{4} = 30 ∴ \frac{a (1 - r^{4})}{1 - r} = 30$ ... $(1)$
$T_{5} + T_{6} + T_{7} + T_{8} = 480$
$∴ {a r}^{4} \frac{(1 - r^{4})}{1 - r} = 480$ ... $(2)$
Dividing $(2)$ by $(1)$ ,
$∴ r^{4} = 16$ or $r^{2} = 4$ or $r = 2, - 2$
Putting in $(1), a = 2, - 6$
The $G . P .^{'} s$ are $a = 2,$ $r = 2$ and $a = - 6, r = - 2$
2nd part :
$\frac{T_{n}}{T_{n}^{'}} = \frac{a r^{n - 1}}{a^{'} {r^{'}}^{n - 1}} = \frac{2}{- 6} {(\frac{2}{- 2})}^{n - 1}$
$= - \frac{1}{3} {(- 1)}^{n - 1} = \frac{1}{3} {(- 1)}^{n}$
$S_{6} = \frac{a (r^{5} - 1)}{r - 1} = \frac{2 (64 - 1)}{2 - 1} = 126$
$∴ S_{6} {- S}_{4} = 126 - 30 = 96$
Similarly, $S_{6} {^{'} - S}_{4}^{'} = 96.$

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