Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

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

Option 1)
$\frac{1}{12}$

Option 2)
$\frac{7}{116}$

Option 3)
$\frac{21}{346}$

Option 4)
$\frac{29}{358}$

Answers (1)

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

$\alpha +\beta =\frac{25}{375}\: \: \: ,\: \: \: \: \alpha \beta =\frac{-2}{375}$

$\Rightarrow \lim_{n\rightarrow \infty }\sum_{r=1}^{n}\alpha ^{r}+\lim_{n\rightarrow \infty }\sum_{r=1}^{n}\beta ^{r}$

$\Rightarrow \left ( \alpha +\alpha ^{2} +\cdots \infty \right )+\left ( \beta +\beta ^{2}+\cdots \infty \right )$

$=\frac{\alpha }{1-\alpha }+\frac{\beta }{1-\beta }$ $\because G.P=\frac{G}{1-r}$

$=\frac{\alpha \left ( 1-\beta \right )+\beta \left ( 1-\alpha \right )}{\left ( 1-\alpha \right )\left ( 1-\beta \right )}$

$=\frac{\alpha +\beta -2\alpha \beta }{1-\alpha -\beta +\alpha \beta }$

$=\frac{\frac{25}{375}-2-\frac{2}{375}}{1-\frac{25}{375}+\frac{-2}{375}}$

$\frac{25+4}{375-25-2}=\frac{29}{375-27}=\frac{29}{348}=\frac{1}{12}$

Option 1)

$\frac{1}{12}$

Option 2)

$\frac{7}{116}$

Option 3)

$\frac{21}{346}$

Option 4)

$\frac{29}{358}$

Posted by

Vakul

View full answer

JEE Main high-scoring chapters and topics

Study 40% syllabus and score up to 100% marks in JEE

Similar Questions

Trending Articles/News

anam-khan

When will NTA issue JEE Main 2026 city slip for over 14.5 lakh? previous years’ trends

anam-khan

Engineering Entrance Exams 2026: JEE Main from January 21; multiple admission routes, shifting career choices

anam-khan

JEE Main 2026: NTA directs non-Aadhaar applicants to submit photo verification certificate by January 7

Latest Question