Question

If $a,b,c$ are in $A.P$. and $a^2,b^2,c^2$ are in $H.P$.,then

• A. $a=b=c$
• B. $2b=3a+c$
• C. $b^2=\sqrt{\frac{ac}{8}}$
• D. None of these

Answer 1

The correct option is A

$a=b=c$

Explanation for correct option:

Progression:

Given, $a,b,c$ are in Arithmetic Progression

$$\begin{gathered} 2b=a+c \\ \Rightarrow b-a=c-b \\ \Rightarrow a-b=b-c \end{gathered}$$

Also, $a^2,b^2\&c^2$ are in ‘Harmonic progression’, then

$\begin{array}{rcl}\frac{1}{b^2}-\frac{1}{a^2}& =& \frac{1}{c^2}-\frac{1}{b^2}\\ \Rightarrow \frac{(a+b)(a-b)}{{\left(ab\right)}^2}& =& \frac{(b-c)(b+c)}{{\left(bc\right)}^2}\\ \Rightarrow \frac{(a+b)(b-c)}{a^2}& =& \frac{(b-c)(b+c)}{c^2}\\ \Rightarrow \frac{(a+b)}{a^2}& =& \frac{(b+c)}{c^2}\\ \Rightarrow c^2(a+b)& =& a^2(b+c)\\ \Rightarrow c^{2}a+c^{2}b& =& a^{2}b+a^{2}c\\ & \Rightarrow & c^{2}a+c^{2}b-a^{2}b-a^{2}c=0\\ \Rightarrow ac(a-c)+b(a^2-c^2)& =& 0\\ \Rightarrow (a-c)(ac+ab+bc)& =& 0\\ \Rightarrow a& =& c,(ac+ab+bc)=0\end{array}$

Put, $a=c$in equation $\left(\mathrm{i}\right)$

$\begin{array}{rcl}2b& =& 2c\\ b& =& c\end{array}$

Therefore, $\mathrm{a}=\mathrm{b}=\mathrm{c}$

Hence, correct option is $\left(A\right)$