JEE Mains · Maths · STD 11 - 8. sequence and series
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 \(8\)
Answer & Solution
Correct Answer
(A) \(\frac{80}{9}\)
Step-by-step Solution
Detailed explanation
\( p x^2+q x-r=0 < \beta \) \( p=A, q=A R, r=A R^2\) \( A x^2+A R x-A R^2=0\) \( x^2+R x-R^2=0 < \beta \) \( \because \frac{1}{\alpha}+\frac{1}{\beta}=\frac{3}{4} \)…
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