Question

If$‘a,b\&c’$ are in Arithmetic Progression and $a^2,b^2,c^2$are in Harmonic Progression, then

• A. $a\ne b\ne c$
• B. $a^2=b^2=\frac{c^2}{2}$
• C. $‘a,b\&c’$ are in G,P
• D. $\frac{-a}{2},b,c$ are in G.P

Answer 1

The correct option is D

$\frac{-a}{2},b,c$ are in G.P

Explanation for correct option:

Step-1: Expressing the given data:

Given that,

$\because a,b,c$ are in Arithmetic Progression

$\Rightarrow$$\begin{array}{rcl}2b& =& a+c\dots \left(i\right)\end{array}$

$\begin{array}{rcl}& & a^2,b^2,c^2\end{array}$ are in harmonic progression

So, $\begin{array}{rcl}& & \frac{1}{a^2},\frac{1}{b^2},\frac{1}{c^2}\end{array}$are in arithmetic progression

Step2. Finding progression.

$\Rightarrow$ $\begin{array}{rcl}\frac{1}{b^2}–\frac{1}{a^2}& =& \frac{1}{c^2}–\frac{1}{b^2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{(a^2-b^2)}{a^2{b}^2}& =& \frac{(b^2-c^2)}{b^2{c}^2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{(a+b)(a-b)}{a^2}& =& \frac{(b-c)(b+c)}{c^2}\end{array}$ $\begin{array}{rcl}& & \left[\because a-b=b-c\right]\end{array}$

$\Rightarrow$ $\begin{array}{rcl}\frac{(a+b)}{a^2}& =& \frac{(b+c)}{c^2}\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{a}^2(b+c)& =& c^2(a+b)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{a}^{2}b+a^{2}c& =& c^{2}a+c^{2}b\end{array}$

$\Rightarrow$ $\begin{array}{rcl}ac(c-a)+b(c-a)(c+a)& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}(c-a)(ab+bc+ca)& =& 0\end{array}$

$\Rightarrow$$\begin{array}{rcl}(c-a)& =& 0or(ab+bc+ca)=0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}c& =& -aor(a+c)b+ca=0\end{array}$

Substitute $\left(i\right)$ in $\begin{array}{rcl}(a+c)b+ca& =& 0\end{array}$

$\begin{array}{rcl}2b^2+ca& =& 0\end{array}$

$\Rightarrow$ $\begin{array}{rcl}{b}^2& =& \frac{-ac}{2}\end{array}$

$\therefore -\frac{a}{2},b,c$are in Geometric Progression

Hence, correct option is $\mathbf{\left(}\mathit{D}\mathbf{\right)}\mathbf{.}$