If $\alpha$ and $\beta$ are the roots of the equation $375 x^{2}-25 x-2=0$, then $\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \alpha^{r}+\lim _{n \rightarrow \infty} \sum_{r=1}^{n} \beta^{r}$ is equal to :

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Sequences and Series

α

and

β

are the roots of the equation

375 x^{2} - 25 x - 2 = 0

, then

lim_{n \to \infty} \sum_{r = 1}^{n} α^{r} + lim_{n \to \infty} \sum_{r = 1}^{n} β^{r}

is equal to :

Answer: (A)

By Vieta's formulas on

375 x^{2} - 25 x - 2 = 0

, the roots satisfy

α + β = \frac{25}{375} = \frac{1}{15}

and

α β = - \frac{2}{375}

. Both roots have magnitude less than

1

, so each geometric series converges to

\sum_{r = 1}^{\infty} α^{r} = \frac{α}{1 - α}

and likewise for

β

. Adding,

\frac{α}{1 - α} + \frac{β}{1 - β} = \frac{α + β - 2 α β}{1 - ( α + β ) + α β}

. The numerator is

\frac{1}{15} + \frac{4}{375} = \frac{29}{375}

and the denominator is

1 - \frac{1}{15} - \frac{2}{375} = \frac{348}{375}

, so the sum is

\frac{29}{348} = \frac{1}{12}

JEE Main · 2019 · 12 Apr 2019 · Shift 1 · Single correct option