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}

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