Question

If $a,b,care\text{in arithmetic progression.}$ and geometric means of $acandab,abandbc,caandcbared,e,f$respectively then $e+f,f+dandd+eare$ in

• A. $G.P$
• B. $A.P$
• C. $H.P$
• D. $A.G.P$

Answer 1

The correct option is C

$H.P$

Explanation for the correct option:

Step-1: Expressing the given that:

Geometric mean of two number is $\sqrt{\mathbf{x}\mathbf{y}}$

$\begin{array}{rcl}\therefore d^2& =& ac.ab\\ d^2& =& a^{2}bc\end{array}$

Geometric Mean of

$\begin{array}{rcl}ab\&bc& =& e\\ \therefore e^2& =& a{b}^{2}c\end{array}$

Geometric Mean of

$\begin{array}{rcl}ca\&cb& =& f\\ \therefore f^2& =& ab{c}^2\end{array}$

Step-2: Finding relation between $e+f,f+d\&d+e$:

Since $a,b,c$are in arithmetic progression.

Therefore,$\mathit{b}\mathbf{=}\frac{\mathbf{a}\mathbf{+}\mathbf{c}}{\mathbf{2}}$

Now, Using

$\begin{array}{rcl}{d}^2+f^2& =& a^{2}bc+ab{c}^2\\ & =& abc(a+c)\mathbf{[}\mathit{b}\mathbf{=}\frac{\mathbf{a}\mathbf{+}\mathbf{c}}{\mathbf{2}}]\\ & =& \frac{(a+c)}{2}ac(a+c)\\ & =& \frac{ac{(a+c)}^2}{2}\mathbf{}\mathbf{}\mathbf{[}\mathit{m}\mathit{u}\mathit{l}\mathit{t}\mathit{i}\mathit{p}\mathit{l}\mathit{y}\mathbf{}\mathit{w}\mathit{i}\mathit{t}\mathit{h}\mathbf{}\frac{\mathbf{2}}{\mathbf{2}}]\\ & =& \frac{2ac{(a+c)}^2}{4}\\ & =& 2ac{\left(\frac{a+c}{2}\right)}^2=2ac{b}^2\\ d^2+f^2& =& 2e^2\end{array}$

So, $\begin{array}{rcl}& & d^2,e^2,f^2\end{array}$ are in Arithmetic progression.

$\Rightarrow$$\begin{array}{rcl}& & d^2+de+ef+df,e^2+de+ef+df,f^2+de+ef+df\end{array}$ are in Arithmetic progression.

$\Rightarrow$$\begin{array}{rcl}& & (d+e)(d+f),(e+d)(e+f),(f+d)(f+e)\end{array}$ are in Arithmetic Progression.

Divide by, $\begin{array}{rcl}& & (d+e)(e+f)(f+d)\end{array}$

$\Rightarrow$$\begin{array}{rcl}& & \frac{1}{e+f},\frac{1}{d+f},\frac{1}{d+e}\end{array}$are in Arithmetic Progression.

So, $\begin{array}{rcl}& & (e+f),(d+f),(d+e)\end{array}$are in Harmonic Progression.

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