The number $a$ , $b$ , $c$ and $d$ are in G.P. Then the value of ${(a - c)}^{2} + {(b - c)}^{2} + {(b - d)}^{2} - {(a - d)}^{2}$ is

0

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1

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2

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Solution

The correct option is A $0$
$b^{2} = a c$ ............(i)

$b d = c^{2}$ ............(ii)

On dividing the two equations,

$\frac{b}{d} = \frac{a}{c}$ ...........(iii)

According to question,

$(a - c)^{2} + (b - c)^{2} + (b - d)^{2} - (a - d)^{2}$

$= a^{2} + c^{2} - 2 a c + b^{2} + c^{2} - 2 b c + b^{2} + d^{2} - 2 b d - a^{2} - d^{2} + 2 a d$
$= 2 (b^{2} - a c) + 2 (c^{2} - b d) - 2 b c + 2 a d = 2 * 0 + 2 * 0 - 2 b c + 2 b c = 0$ [From (i) and (ii)]

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a, b, c

d

(a - c)^{2} + (b - c)^{2} + (b - d)^{2} - (a - d)^{2}

a, b, c, d

G . P

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

If a,b,c,d are in G.P., then ${(a + b)}^{2}$ , ${(b + c)}^{2}$ , ${(c + d)}^{2}$ are in:

(a + b)^{2}, (b + c)^{2}, (c + d)^{2}

\frac{1}{a^{2} + b^{2}}, \frac{1}{b^{2} + c^{2}}, \frac{1}{c^{2} + d^{2}}

a^{2} - b^{2}, b^{2} - c^{2}, c^{2} - d^{2}

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