if the sum S_4 of the first 4 terms of a G P is 24 and the sum S_6 of the first 6 terms is 30 find the first term and common ratio..


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Question Number 34169 by Rio Mike last updated on 01/May/18

$i f t h e s u m S_{4} o f t h e f i r s t 4 t e r m s$ $o f a G P i s 24 a n d t h e s u m S_{6} o f$ $t h e f i r s t 6 t e r m s i s 30 f i n d t h e$ $f i r s t t e r m a n d c o m m o n r a t i o . .$
	Answered by MJS last updated on 02/May/18

	$. . . really a GP ? You {won}^{'} t like the solution . . .$ $a_{0} (1 + r + r^{2} + r^{3}) = 24 \Rightarrow a_{0} = \frac{24}{1 + r + r^{2} + r^{3}}$ $a_{0} (1 + r + r^{2} + r^{3} + r^{4} + r^{5}) = 30$ $24 \frac{1 + r + r^{2} + r^{3} + r^{4} + r^{5}}{1 + r + r^{2} + r^{3}} = 30$ $24 \frac{1 + r^{2} + r^{4}}{1 + r^{2}} = 30$ $r^{4} - \frac{1}{4} r^{2} - \frac{1}{4} = 0$ $r^{2} = \frac{1}{8} \pm \frac{\sqrt{17}}{8}$ $r^{2} > 0 \Rightarrow r = \pm \frac{\sqrt{1 + \sqrt{17}}}{\sqrt{8}} = - \frac{\sqrt{2 + 2 \sqrt{17}}}{4} \lor \frac{\sqrt{2 + 2 \sqrt{17}}}{4}$ $a_{0} = \frac{3}{8} (4 + \sqrt{2 + 2 \sqrt{17}}) (23 + \sqrt{17}) \lor \frac{3}{8} (4 - \sqrt{2 + 2 \sqrt{17}}) (23 + \sqrt{17})$ $(r \approx - . 800243 \land a_{0} \approx 73.2423) \lor$ $(r \approx . 800243 \land a_{0} \approx 8.12706)$
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