a,b,c is a geometric progression such that a+b+c=26 a^2 +b^2 +c^2 =364 find a,b,c


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Question Number 64519 by necx1 last updated on 18/Jul/19

$a, b, c i s a g e o m e t r i c p r o g r e s s i o n s u c h$ $t h a t$ $a + b + c = 26$ $a^{2} + b^{2} + c^{2} = 364$ $f i n d a, b, c$
	Answered by MJS last updated on 19/Jul/19

	$a + a r + {a r}^{2} = 26 \Rightarrow a = \frac{26}{1 + r + r^{2}}$ $a^{2} + a^{2} r^{2} + a^{2} r^{4} = 364 \Rightarrow a^{2} = \frac{364}{1 + r^{2} + r^{4}}$ $\frac{26^{2}}{{(1 + r + r^{2})}^{2}} = \frac{364}{1 + r^{2} + r^{4}}$ $13 (1 + r^{2} + r^{4}) = 7 {(1 + r + r^{2})}^{2}$ $6 r^{4} - 14 r^{3} - 8 r^{2} - 14 r + 6 = 0$ $6 (r - 3) (r - \frac{1}{3}) (x^{2} + x + 1) = 0$ $\Rightarrow r = 3 \lor r = \frac{1}{3}$ $r = 3 \Rightarrow a = 2 b = 6 c = 18$ $r = \frac{1}{3} \Rightarrow a = 18 b = 6 c = 2$
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