If $\alpha $ and $\beta $ the roots of ${x^2} - x - 1 = 0\;\,{\rm{and}}\;{A_n} = {\alpha ^n} + {\beta ^n},\;\,{\rm{then\;AM}}\;{\rm{of}}\;{A_{n - 1}}\,\,{\rm{and}}\;{A_n}$ is

JEE MATHEMATICS MCQ: QUADRATIC EQUATIONS

1 Easy

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Application +4 | -1

\alpha

and

\beta

the roots of

{x^2} - x - 1 = 0\;\,\mathrm{and}\;{A_n} = {\alpha ^n} + {\beta ^n},\;\,\mathrm{then\;AM}\;\mathrm{of}\;{A_{n - 1}}\,\,\mathrm{and}\;{A_n}

View answer and explanation

Correct option: B

\frac{{{A_{n + 1\;}}}}{2}

Option B (

\frac{{{{A_{n + 1\;}}}}}{{{{2}}}}

) is correct because it is derived by using the properties of roots of the given quadratic equation and simplifying the expression for the arithmetic mean.

More Questions from QUADRATIC EQUATIONS

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The solution of the equation

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

is If

\alpha ,\beta

are the roots of

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

and

\alpha < 1 < \beta ,

then If

\alpha ,\;\beta

be the roots of the equation

{x^2} - px + q = 0

and

{\alpha _1},{\beta _1}

be the roots of the equation

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

then the equation whose roots are

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

is If

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

for all values of

x

, then

a

lies in the interval If

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

and

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

and

\;a \ne 0,

then Root(s) of the equation

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

belonging to the domain of definition of the function

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

is (are)