The sum of the squares of three distinct real numbers which are in GP is S^2 . If their sum is α S, then


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Question Number 7525 by Little last updated on 02/Sep/16

$The sum of the squares of three distinct$ $real numbers which are in GP is S^{2} . If$ $their sum is α S, then$
Commented by Rasheed Soomro last updated on 04/Sep/16

$A s s u m e d t h a t α i s r e q u i r e d .$ $L e t a i s t h e i n i t i a l r e a l n u m b e r a n d r i s$ $c o m m o n r a t i o .$ $S o t h r e e d i s t i n c t r e a l n u m b e r s a r e$ $a, a r, {a r}^{2}$ $S u m o f s q u a r e s : a^{2} + a^{2} r^{2} + a^{2} r^{4} = S^{2} . . . . . . . . . . . (i)$ $S u m o f r e a l n u m b e r s : a + a r + {a r}^{2} = α S$ $O r S = \frac{a (1 + r + r^{2})}{α} . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . (i i)$ $F r o m (i) & (i i)$ $a^{2} (1 + r^{2} + r^{4}) = {(\frac{a (1 + r + r^{2})}{α})}^{2}$ $1 + r^{2} + r^{4} = \frac{{(1 + r + r^{2})}^{2}}{α^{2}}$ $α^{2} = \frac{{(1 + r + r^{2})}^{2}}{1 + r^{2} + r^{4}}$
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