If (a - b), (b - c), (c - a) are in G.P. then prove that $(a + b + c)^{2} = 3 (a b + b c + c a)$

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Solution

$(a - b), (b - c), (c - a)$ are in G.P. $(b + c)^{2} = (a - b) (c - a)$

$b^{2} + c^{2} - 2 b c = a c - a^{2} - b c + a b$

$b^{2} + c^{2} + a^{2} = a c + b c + a b . . . . . (i)$

Now,

$(a + b + c)^{2} = a^{2} + b^{2} + c^{2} + 2 a b + 2 b c + 2 c a$

$= a c + b c + a b + 2 a b + 2 b c + 2 c a$

Using equation (i)

$= 3 a b + 3 b c + 3 c a$

$(a + b + c)^{2} = 3 (a b + b c + c a)$

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If a, b, c are in G.P. then prove that : $\frac{a^{2} + a b + b^{2}}{b c + c a + a b} = \frac{b + a}{c + b}$

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