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8. Sequences and Series
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Let $\alpha$ and $\beta$ be the roots of the equation $\mathrm{px}^2+\mathrm{qx}-$ $r=0$, where $p \neq 0$. If $p, q$ and $r$ be the consecutive terms of a non-constant G.P and $\frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4}$, then the value of $(\alpha-\beta)^2$ is :
A
$\frac{80}{9}$
B
$9$
C
$\frac{20}{3}$
D
$8$
(JEE MAIN-2024)
Solution
$ p x^2+q x-r=0 < \beta $
$ p=A, q=A R, r=A R^2$
$ A x^2+A R x-A R^2=0$
$ x^2+R x-R^2=0 < \beta $
$ \because \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} $
$ \therefore \frac{\alpha+\beta}{\alpha \beta}=\frac{3}{4} \Rightarrow \frac{-R}{-R^2}=\frac{3}{4} \Rightarrow R=\frac{4}{3} $
$ (\alpha-\beta)^2=(\alpha+\beta)^2-4 \alpha \beta=R^2-4\left(-R^2\right)=5\left(\frac{16}{9}\right) $
$ =80 / 9$
Standard 11
Mathematics
