Question

If $\alpha ,\beta$ are the roots of $a{x}^2-bx+c=0$ and $\gamma ,\delta$ are the roots of $p{x}^2-qx+r=0$ and if $\alpha ,\beta ,\gamma ,\delta$ are in GP, then the common ratio is equal to

• A. ${\left(\frac{ar}{cp}\right)}^{\frac{1}{4}}$
• B. ${\left(\frac{ar}{cp}\right)}^{\frac{1}{8}}$
• C. ${\left(\frac{ap}{cr}\right)}^{\frac{1}{4}}$
• D. ${\left(\frac{ar}{cp}\right)}^{\frac{-1}{4}}$

Answer 1

The correct option is A

${\left(\frac{ar}{cp}\right)}^{\frac{1}{4}}$

Explanation for the correct option:

Step 1. Find the common ratio:

Given, $\alpha ,\beta ,\gamma ,\delta$ are in GP.

Let $\alpha =A,\beta =Ar,\gamma =A{r}^2,\delta =A{r}^3$, where $A$ is the first term of a GP and $r$ is the common ratio.

Also, $\alpha ,\beta$ are the roots of $a{x}^2-bx+c=0$

$\therefore$Product of roots, $\alpha \beta =A^{2}r$

$\Rightarrow$ $\alpha \beta =\frac{c}{a}$ …(i)

$\gamma ,\delta$ are the roots of $p{x}^2-qx+r=0$

$\therefore$Product of roots, $\gamma \delta =A^2{r}^5$

$\Rightarrow$ $\gamma \delta =\frac{r}{p}$…(ii)

Step 2. By Dividing equation (ii) by (i), we get

$\begin{array}{rcl}\frac{\gamma \delta }{\alpha \beta }& =& \frac{A^2{r}^5}{A^{2}r}\\ & =& \frac{{\displaystyle \frac{r}{p}}}{{\displaystyle \frac{c}{a}}}\end{array}$

$\Rightarrow$$r^4=\frac{ar}{cp}$

$\mathbf{\therefore }\mathit{r}\mathbf{}\mathbf{=}\mathbf{}{\left(\frac{ar}{cp}\right)}^{\frac{\mathbf{1}}{\mathbf{4}}}$

Hence, Option ‘A’ is Correct.