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Let p ,q be integers and let alpha,beta ...

Let `p ,q` be integers and let `alpha,beta` be the roots of the equation, `x^2-x-1=0,` where `alpha!=beta` . For `n=0,1,2, ,l e ta_n=palpha^n+qbeta^ndot` FACT : If `aa n db` are rational number and `a+bsqrt(5)=0,t h e na=0=bdot` If `a_4=28 ,t h e np+2q=` 7 (b) 21 (c) 14 (d) 12

A

`a_(11) - a_(10)`

B

`a_(11) + a_(10)`

C

`2a_(11) + a_(10)`

D

` a_(11) +2a_(10)`

Text Solution

Verified by Experts

The correct Answer is:
2
Since ` alpha, beta ` be the roots of the equation, ` x^(2) - x - 1= 0` ,
` alpha ^(2) - alpha - 1 = 0 and beta ^(2) - beta - 1 = 0` …(i)
`a_(n) + a_(n-1) = palpha ^(n) + q beta^(n) + palpha ^(n-1) + q beta ^(n-1)`
` = p alpha ^(n) + p alpha ^(n-1) + q beta^(n) + q beta^(n-1)`
` = p alpha ^(n-1) + (alpha +1)+ q beta^(n-1) + (beta+ 1)`
` = p alpha ^(n-1) . alpha ^(2) + q beta^(n-1) . beta^(2)` [Ising (i)]
` p lapha ^(n+1) + q beta ^(n+1)`
` a_(n+ 1)`
So, ` a_(12) = a_(11) + a_(10)`
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