Question

If $a^2,b^2,c^2$ are in $AP$, then which of the following are in $AP$?

• A. $(b+c),(c+a),(a+b)$
• B. $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$
• C. $\frac{b+c}{a},\frac{c+a}{b},\frac{a+b}{c}$
• D. $\frac{1}{a^2},\frac{1}{a^2},\frac{1}{c^2}$

Answer 1

The correct option is B $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$

$2b^2=a^2+c^2$

$b^2+b^2=a^2+c^2$

$b^2-a^2=c^2-b^2$

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

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

Dividing both sides by $(c+a)$

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

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

$\frac{1}{(c+a)}-\frac{1}{b+c}=\frac{1}{(a+b)}-\frac{1}{c+a}$

$\frac{2}{(c+a)}=\frac{1}{(a+b)}-\frac{1}{(b+c)}$

Therefore $\frac{1}{(a+b)},\frac{1}{(b+c)},\frac{1}{(c+a)}$ are in AP.