Question

$\frac{1}{a},\frac{1}{b},\frac{1}{c}$ are in $H.P$, then the following are also in A.P.
$b+c,c+a,a+b$

If it is true, then answer will be 1 otherwise 0.

Answer 1

We have,

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

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

So,

$2b=a+c$ $............\left(1\right)$

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

$b+b+c+a=c+a+c+a$

$b+c+a+b=2(c+a)$

$2(c+a)=b+c+a+b$

So,

$b+c,c+a,a+b$ are in $A.P$

Hence, this is the answer.