Given a,b,c,d are in G.P then $(a - d)^{2} = (b - c)^{2} + (c - a)^{2} + (d - b)^{2}$ .If you think this is true then write 1 otherwise write 0 ?

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Solution

R.H.S. $= (b - c)^{2} + (c - a)^{2} + (d - b)^{2}$
$= 2 (b^{2} - a c) + 2 (c^{2} - b d) + (d^{2} + a^{2} - 2 b c)$
Now given $a, b, c, d$ are in in G.P.,

so we have $\frac{b}{a} = \frac{c}{b} = \frac{d}{c} \Rightarrow b^{2} = a c, c^{2} = b d$ and $b c = a d$
$∴ R . H . S . = 0 + 0 + (d^{2} + a^{2} - 2 a d) = (d - a)^{2}$
Thus answer is 1.

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