If $\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b}$ be consecutive terms of an AP then $(b - c)^{2}, (c - a)^{2}, (a - b)^{2}$ will be in ?

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None of these

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Solution

The correct option is B AP
If $a, b, c$ is in $A P$ then we know that,
$b - a = c - b = d$ where $d$ is the common difference
$\Rightarrow \frac{1}{c - a} - \frac{1}{b - c} = \frac{1}{a - b} - \frac{1}{c - a} (\frac{1}{b - c}, \frac{1}{c - a}, \frac{1}{a - b} a r e i n A P) \Rightarrow \frac{b - c - c + a}{(c - a) (b - c)} = \frac{c - a - a - b}{(a - b) (c - a)} \Rightarrow a^{2} - 2 a c - b^{2} + 2 b c = b^{2} - 2 a b - c^{2} + 2 a c \Rightarrow a^{2} - 2 a c + c^{2} - c^{2} - b^{2} + 2 b c = b^{2} - 2 a b + a^{2} - a^{2} - c^{2} + 2 a c \Rightarrow (c^{2} - 2 a c + a^{2}) - (b^{2} - 2 b c + c^{2}) = (a^{2} - 2 a b + b^{2}) - (c^{2} - 2 a c + a^{2}) \Rightarrow {(c - a)}^{2} - {(b - c)}^{2} = {(a - b)}^{2} - {(c - a)}^{2}$
$∴ {(b - c)}^{2}, {(c - a)}^{2}, {(a - b)}^{2} a r e i n A P$

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If $\frac{1}{b - c}$ , $\frac{1}{c - a}$ , $\frac{1}{a - b}$ be consecutive terms of an A.P., then $(b - c)^{2}, (c - a)^{2}, (a - b)^{2}$ will be in ___.

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