Question

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 ___.

• A. G.P
• B. A.P
• C. H.P
• D. None of these

Answer 1

The correct option is B

A.P

If $(b-c{)}^2,(c-a{)}^2,(a-b{)}^2$ are in A.P.

Then we have $(c-a{)}^2-(b-c{)}^2$ = $(a-b{)}^2-(c-a{)}^2$

$\Rightarrow (b-a)(2c-a-b)$ = (c - b)(2a - b - c) ......(i)

Also if $\frac{1}{b-c},\frac{1}{c-a},\frac{1}{a-b}$ are in A.P.

Then $\frac{1}{c-a}-\frac{1}{b-c}$ = $\frac{1}{a-b}-\frac{1}{c-a}$

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

$\Rightarrow (a-b)(b+a-2c)$ = $(b-c)(c+b-2a)$

$\Rightarrow (b-a)(2c-a-b)$ = $(c-b)(2a-b-c)$

Which is true by virtue of (i).