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

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Medium

1 of 1

Application +4 | -1

\alpha

and

\beta

are the roots of the equation

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

, then

\lim\limits_{n \to \infty } \,\sum\limits_{r = 1}^n {{\alpha ^r} + } \lim\limits_{n \to \infty } \,\,\sum\limits_{r = 1}^n {{\beta ^r}}

is equal to

View answer and explanation

Correct option: C

\frac{{1}}{{12}}

Option C (

\frac{{1}}{{12}}

) is correct because it is the result of summing the infinite geometric series formed by the roots of the given quadratic equation.

Practice All Chapter MCQs →

\log _7 \log _5(\sqrt{x+5}+\sqrt{x})=0

\alpha ,\beta

{x^2} - 3x + a = 0,a \in R

\alpha < 1 < \beta ,

\alpha ,\;\beta

{x^2} - px + q = 0

{\alpha _1},{\beta _1}

{x^2} - qx + p = 0,

\frac{1}{{{\alpha _1}\beta }} + \frac{1}{{\alpha {\beta _1}}}\mathrm{and}\;\frac{1}{{\alpha {\alpha _1}}} + \frac{1}{{\beta {\beta _1}}},

{x^2} + 4ax + 2 > 0

x

a

f\left( x \right) = a{x^2} + bx + c\;

\;f\left( { - 1} \right) < 1,f\left( 1 \right) > - 1,f\left( 3 \right) < - 4\;

\;a \ne 0,

9{x^2} - 18\left| x \right| + 5 = 0

f\left( x \right) = \log \left( {{x^2} - x - 2} \right)