$$\begin{gathered} If\alpha and\beta aretherootsoftheequation375x^2-25x-2=0,then \\ \underset{n\to \infty }{\lim }\sum _{r=1}^n{\alpha }^r+\underset{n\to \infty }{\lim }\sum _{r=1}^n{\beta }^{r}isequalto \end{gathered}$$

$\begin{array}{l}If\alpha ,\beta arerootsof375x^2-25x-2=0\\ Then\alpha +\beta =\frac{25}{375},\alpha \beta =\frac{-2}{375}\\ Now\underset{n\to \infty }{lt}\sum _{r=1}^n{\alpha }^2+\underset{n\to \infty }{lt}\sum _{r=1}^n{\beta }^2\\ =(\alpha +{\alpha }^2+{\alpha }^3+---\infty )+(\beta +{\beta }^2+{\beta }^3+---\infty )\\ =\frac{\alpha }{1-\alpha }+\frac{\beta }{1-\beta }\\ =\frac{\alpha (1-\beta )+\beta (1-\alpha )}{(1-\alpha )(1-\beta )}\\ =\frac{(\alpha +\beta )-2\alpha \beta }{1-(\alpha +\beta )+\alpha \beta }\\ =\frac{25+4}{375-25-2}\\ =\frac{29}{348}\\ =\frac{1}{12}\end{array}$