QJEEM15Ma929e: JEE Main 2015 Mathematics - Class 11 JEEM15Ma929e Solution: Let and be the roots of equation . If for then the value of is…

Question

Verified Solution • 2025–26 Session moderate

Let

\alpha

and

\beta

be the roots of equation

{x^2} - 6x - 2 = 0

. If

{a_n} = {\alpha ^n} - {\beta ^n},

for

n \ge 1,

then the value of

{{{a_{10}} - 2{a_8}} \over {2{a_9}}}

is equal to :

Accepted Answer

Given equation, x² - 6x - 2 = 0

Roots are

\alpha

and

\beta

.

So,

\alpha + \beta = 6

and

\alpha \beta = - 2

In the question given,

{a_n} = {\alpha ^n} - {\beta ^n}

\therefore

{a_8} = {\alpha ^8} - {\beta ^8}

and

{a_9} = {\alpha ^9} - {\beta ^9}

and

{a_{{10}}} = {\alpha ^{{10}}} - {\beta ^{10}}

Now, the given equation

{{{a_{10}} - 2{a_8}} \over {2{a_9}}}

=

{{{\alpha ^{10}} - {\beta ^{10}} - 2\left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}

=

{{{\alpha ^{10}} - {\beta ^{10}} + \alpha \beta \left( {{\alpha ^8} - {\beta ^8}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}

(as

\alpha \beta = - 2

)

=

{{{\alpha ^{10}} - {\beta ^{10}} + {\alpha ^9}\beta - \alpha {\beta ^9}} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}

=

{{{\alpha ^9}\left( {\alpha + \beta } \right) - {\beta ^9}\left( {\alpha + \beta } \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}

=

{{\left( {\alpha + \beta } \right)\left( {{\alpha ^9} - {\beta ^9}} \right)} \over {2\left( {{\alpha ^9} - {\beta ^9}} \right)}}

=

{{\left( {\alpha + \beta } \right)} \over 2}

=

{6 \over 2}

(as

{ {\alpha + \beta } }

= 6)

= 3

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