JEE Main 2019 — Mathematics Algebra

For given quadratic $375{x}^{2}-25x-2=0,$375x2−25x−2=0, its roots are,

$$\alpha ,\beta =\frac{25\pm \sqrt{{25}^{2}+2\times 4\times 375}}{2\times 375}\Rightarrow |\alpha |<1,|\beta |<1$$α,β=2×37525±252+2×4×375​​⇒∣α∣<1,∣β∣<1

Also, $\alpha +\beta =\frac{25}{375},$α+β=37525​, $\alpha \beta =\frac{-2}{375}$αβ=375−2​

Now $\displaystyle \underset{n\rightarrow \infty }{lim}\sum _{r=1}^{n}{\alpha }^{r}+\underset{n\rightarrow \infty }{lim}\sum _{r=1}^{n}{\beta }^{r}$n→∞lim​r=1∑n​αr+n→∞lim​r=1∑n​βr

$=\frac{\alpha }{1-\alpha }+\frac{\beta }{1-\beta }$=1−αα​+1−ββ​ $[\begin{matrix}a+ar+a{r}^{2}+..... \mathrm{infinite}\mathrm{terms} \\ =\frac{a}{1-r} \mathrm{if} |r|<1\end{matrix}]$[a+ar+ar2+.....infiniteterms=1−ra​if∣r∣<1​]

$$=\frac{(\alpha +\beta )-2\alpha \beta }{1-(\alpha +\beta )+\alpha \beta }=\frac{\frac{25}{375}+\frac{4}{375}}{1-\frac{25}{375}-\frac{2}{375}}=\frac{29}{348}=\frac{1}{12}$$=1−(α+β)+αβ(α+β)−2αβ​=1−37525​−3752​37525​+3754​​=34829​=121​