JEE PYQ: Quadratic Equation Question 40
Question 40 - 2019 (12 Apr Shift 1)
If $\alpha$ and $\beta$ are the roots of the equation $375x^2 - 25x - 2 = 0$, then $\lim_{n\to\infty}\sum_{r=1}^{n}\alpha^r + \lim_{n\to\infty}\sum_{r=1}^{n}\beta^r$ is equal to:
(1) $\frac{21}{346}$
(2) $\frac{29}{358}$
(3) $\frac{1}{12}$
(4) $\frac{7}{116}$
Show Answer
Answer: (3)
Solution
Given equation is, $375x^2 - 25x - 2 = 0$
Sum and product of the roots are,
$\alpha + \beta = \frac{25}{375}$ and $\alpha\beta = \frac{-2}{375}$
$\lim_{n\to\infty}\sum (\alpha^r + \beta^r)$
$= (\alpha + \alpha^2 + \alpha^3 + \ldots) + (\beta + \beta^2 + \beta^3 + \ldots + \infty)$
$= \frac{\alpha}{1-\alpha} + \frac{\beta}{1-\beta} = \frac{\alpha + \beta - 2\alpha\beta}{1 - (\alpha + \beta) + \alpha\beta}$
$= \frac{\frac{25}{375} + \frac{4}{375}}{1 - \frac{25}{375} - \frac{2}{375}} = \frac{29}{375 - 25 - 2} = \frac{29}{348} = \frac{1}{12}$
BETA
