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-

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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