Let $a, b$ be two non-zero real numbers. If $p$ and $r$ are the roots of the equation $x^{2}-8 \mathrm{a} x+2 \mathrm{a}=0$ and $\mathrm{q}$ and s are the roots of the equation $x^{2}+12 \mathrm{~b} x+6 \mathrm{~b}=0$, such that $$\frac{1}{\mathrm{p}}, \frac{1}{\mathrm{q}}, \frac{1}{\mathrm{r}}, \frac{1}{\mathrm{~s}}$$ are in A.P., then $\mathrm{a}^{-1}-\mathrm{b}^{-1}$ is equal to _____________.
Answer (integer)
38
Solution
$\because$ Roots of $2 a x^{2}-8 a x+1=0$ are $\frac{1}{p}$ and $\frac{1}{r}$ and roots of $6 b x^{2}+12 b x+1=0$ are $\frac{1}{q}$ and $\frac{1}{s}$.
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Let $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ as $\alpha-3 \beta, \alpha-\beta, \alpha+\beta, \alpha+3 \beta$
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So sum of roots $2 \alpha-2 \beta=4$ and $2 \alpha+2 \beta=-2$
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Clearly $\alpha=\frac{1}{2}$ and $\beta=-\frac{3}{2}$ <br/><br/>Now product of roots, $\frac{1}{p} \cdot \frac{1}{r}=\frac{1}{2 a}=-5 \Rightarrow \frac{1}{a}=-10$<br/><br/> and $\frac{1}{q} \cdot \frac{1}{x}=\frac{1}{6 b}=-8 \Rightarrow \frac{1}{b}=-48$
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So, $\frac{1}{a}-\frac{1}{b}=38$
About this question
Subject: Mathematics · Chapter: Sequences and Series · Topic: Arithmetic Progression
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