Discussion Forum

1)

Let p,q be integers and let α ,β be the roots of the equation $x^{2}-x-1=0$  where α ≠β, For n=0,1,2...... Let  $a_{n}=p\alpha^{n}+q\beta^{n}$  ( If a and b are rational numbers and  $a+b\sqrt{5}=0$, then a=0=b)

If a₂₄=28   , then p+2q=

Answer:

Option D

Explanation:

$\alpha =\frac{1+\sqrt{5}}{2}$

$\beta =\frac{1-\sqrt{5}}{2}$

$a_{4}=a_{3}+a_{2}$

 $=2a_{2}+a_{1}$

 $=3a_{1}+2a_{0}$

$28=p(3\alpha+2)+q(3\beta+2)$

$28=p+q(\frac{3}{2}+2)+p-q(\frac{3\sqrt{5}}{2})$

$\therefore$ p-q=0

and      $(p+q)\times\frac{7}{2}=28$

$\Rightarrow$   p+q=8

$\Rightarrow$   p=q=4

    p+2q=12