Question

If $a^2,b^2,c^2$ are in A.P., then the following are also in A.P.
$\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$.

• A. True
• B. False

Answer 1

The correct option is A True

We have,

$a^2,b^2,c^2$ are in $A.P$

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

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

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

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

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

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

$\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ are in $A.P$

$\frac{a+b+c}{b+c},\frac{a+b+c}{c+a},\frac{a+b+c}{a+b}$ are in $A.P$

$\frac{a}{b+c}+1,\frac{b}{c+a}+1,\frac{c}{a+b}+1$ are in $A.P$

$\frac{a}{b+c},\frac{b}{c+a},\frac{c}{a+b}$ are in $A.P$

Hence, this is the answer.