Question

If $a,b,c,d$ are real numbers and $a^2,b^2,c^2$ are in $AP$ then which of the following combinations are in $AP$.

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

Answer 1

The correct option is A $\frac{1}{(b+c)},\frac{1}{(a+b)},\frac{1}{(c+a)}$

$\begin{array}{c}{a}^2,b^2,c^{2}areinA.P\\ \therefore b^2-a^2=c^2-b^2\\ b^2+b^2=a^2+c^2.............\left(1\right)\end{array}$

Now, we need to prove $\frac{1}{b+c},\frac{1}{c+a},\frac{1}{a+b}$ are in $A.P$

$\therefore y$ they are in $A.P,\frac{1}{c+a}-\frac{1}{b+c}=\frac{1}{a+b}=\frac{1}{c+a}$

$\begin{array}{c}\frac{\left(b+c\right)-\left(a+c\right)}{\left(a+c\right)\left(b+c\right)}=\frac{\left(a+c\right)-\left(a+b\right)}{\left(a+b\right)\left(c+a\right)}\\ \frac{b-a}{\left(a+c\right)\left(b+c\right)}=\frac{c-b}{\left(a+b\right)\left(c+a\right)}\\ \frac{-\left(a-b\right)}{\left(a+c\right)\left(b+c\right)}=\frac{-\left(b-c\right)}{\left(a+b\right)\left(c+a\right)}\\ cross-multiplying,weget\\ \left(a-b\right)\left(a+b\right)\left(a+c\right)=\left(b-c\right)\left(b+c\right)\left(a+c\right)\\ \left(a-b\right)\left(a+b\right)=\left(b-c\right)\left(b+c\right)\\ a^2-b^2=b^2-c^2\\ a^2-b^2=2b^2\end{array}$

Which satisfies equation (1) , Hence

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