If a,b and c are in arithmetic progression and $a^{2}, b^{2}$ and $c^{2}$ are in Harmonic progression, then prove that either a = b = c or a, b, and $\frac{- c}{2}$ are in Geometric progression

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Solution

Given that $a, b, c$ are in A. P.

$\Rightarrow 2 b = a + c$ ……. (1)

And $a^{2}, b^{2}, c^{2}$ are in H. P.

$\frac{1}{b^{2}} - \frac{1}{d^{2}} = \frac{1}{c^{2}} - \frac{1}{b^{2}}$

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

$a c^{2} + b c^{2} = a^{2} b + a^{2} c [∵ a - b = b - c]$

$a c (c - a) + b (c - a) (c + a) = 0$

$(c - a) (a b + b c + c a) = 0$

either $c - a = 0 o r a b + b c + c a = 0$

either $c = a$ or $(a + c) b + c a = 0$ and then form (i) $2 b^{2} + c a = 0$

Either $a = b = c$ or $b^{2} = a \frac{- c}{2}$

i.e. $a, b, \frac{- c}{2}$ are in G. P. Hence Proved

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Similar questions

Let a, b, c are non-zero real numbers such that \frac{1}{a}, \frac{1}{b}, \frac{1}{c} are in arithmetic progression and a, b, -2c are in geometric progression, then which of the following statement(s) must be true?

a^{2}

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Δ A B C

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cot A, cot B, cot C

a x^{2} - 2 b x + c = 0

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