a, b, d are gp. such that a, b, c are real. if a + b + c = 26 and a^2 + b^2 + c^2 = 364, find b ?


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Question Number 55415 by Tawa1 last updated on 23/Feb/19

$a, b, d are gp . such that a, b, c are real . if$ $a + b + c = 26 and a^{2} + b^{2} + c^{2} = 364, find b ?$
	Answered by tanmay.chaudhury50@gmail.com last updated on 23/Feb/19

	$a = a$ $b = a r$ $c = {a r}^{2}$ $a + a r + {a r}^{2} = 26 \to a (1 + r + r^{2}) = 26$ $a^{2} + a^{2} r^{2} + a^{2} r^{4} = 364 \to a^{2} (1 + r^{2} + r^{4}) = 364$ $r^{4} + r^{2} + 1$ $r^{4} + 2 r^{2} + 1 - r^{2}$ ${(r^{2} + 1)}^{2} - {(r)}^{2}$ $(r^{2} + r + 1) (r^{2} - r + 1)$ $a^{2} (r^{2} + r + 1) (r^{2} - r + 1) = 364$ $n o w$ $a (r^{2} - r + 1) = \frac{364}{26} = 14$ $\frac{a (r^{2} + r + 1)}{a (r^{2} - r + 1)} = \frac{26}{14}$ $13 r^{2} - 13 r + 13 = 7 r^{2} + 7 r + 7$ $6 r^{2} - 20 r + 6 = 0$ $3 r^{2} - 10 r + 3 = 0$ $3 r^{2} - 9 r - r + 3 = 0$ $3 r (r - 3) - 1 (r - 3) = 0$ $(r - 3) (3 r - 1) = 0$ $s o r = 3 o r \frac{1}{3}$ $a (r^{2} + r + 1) = 26$ $a (9 + 3 + 1) = 26$ $a = 2 s o$ $b = 2 \times 3 = 6$ $a (\frac{1}{9} + \frac{1}{3} + 1) = 26$ $a (\frac{1 + 3 + 9}{9}) = 26$ $a = \frac{26 \times 9}{13} = 18$ $s o b = a r = 18 \times \frac{1}{3} = 6$
	Commented by Tawa1 last updated on 23/Feb/19

	$God bless you sir . I appreciate your effort$
	Answered by peter frank last updated on 24/Feb/19

	$\frac{b}{a} = \frac{c}{b}$ $b^{2} = ac$ $a^{2} + b^{2} + c^{2} = 364$ $a + b + c = 26$ $a + c = 26 - b$ ${(a + c)}^{2} - 2 ac + b^{2} = 364$ $b^{2} = ac$ ${(a + c)}^{2} - 2 {(b)}^{2} + b^{2} = 364$ ${(a + c)}^{2} - b^{2} = 364$ $(a + c + b) (a + c - b) = 364$ $(26 - b + b) (26 - b - b) = 364$ $26 (26 - 2 b) = 364$ $b = 6$
	Commented by Tawa1 last updated on 24/Feb/19

	$God bless you sir$
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