Let p , q be integers and let be the roots of the equation x 2 - x - 1 = 0 , where . For n = 0 , 1 , 2 , … , let a n = p n + q n . FACT: If a and b are rational numbers and a + b 5 = 0 , then a = 0 = b . Q. If a 4 = 28 , then p

Question

Let $p, q$ be integers and let $α, β$ be the roots of the equation $x^{2} - x - 1 = 0,$ where $α \neq β .$ For $n = 0, 1, 2, \dots,$ let $a_{n} = p α^{n} + q β^{n} .$

FACT: If $a$ and $b$ are rational numbers and $a + b \sqrt{5} = 0,$ then $a = 0 = b .$ [2017]

Q. $If a_{4} = 28, then p + 2 q =$

Smart Achievers · Accepted Answer

(4)

$α = \frac{1 + \sqrt{5}}{2}, β = \frac{1 - \sqrt{5}}{2}$

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

$= 2 a_{2} + a_{1}$

$= 3 a_{1} + 2 a_{0}$

$b$

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

$∴$ $p - q = 0 and (p + q) \cdot \frac{7}{2} = 28$

$\Rightarrow p + q = 8 \Rightarrow p = q = 4$

$∴$ $p + 2 q = 12$

Solution

1 21

2 14

3 7

4 12

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Solution

Q. Let p,q be integers and let α,β be the roots of the equation x2-x-1=0, where α≠β. For n=0,1,2,…, let an=pαn+qβn. FACT: If a and b are rational numbers and a+b5=0, then a=0=b. [2017] Q. If a4=28, then p+2q=

1 21

2 14

3 7

4 12

Ans. (4) α=1+52, β=1-52 a4=a3+a2 =2a2+a1 =3a1+2a0 28=p(3α+2)+q(3β+2) 28=(p+q)(32+2)+(p-q)(352) ∴ p-q=0 and (p+q)·72=28 ⇒p+q=8⇒p=q=4 ∴ p+2q=12

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