Let $p, q$ be integers and let $\alpha, \beta$ be the roots of the equation $x^{2}-x-1=0$, where $\alpha \neq \beta$. For $n=0,1,2, \ldots$, let $a_{n}=p \alpha^{n}+q \beta^{n}$. FACT: If $a$ and $b$ are rational numbers and $a+b \sqrt{5}=0$, then $a=0=b$. $a_{12}=$

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Sequences and Series

Passage

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 5 = 0

, then

a = 0 = b

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 5 = 0

, then

a = 0 = b

a_{12} =

Answer: (B)

Since

α

and

β

are roots of

x^{2} - x - 1 = 0

, we have:

α^{2} = α + 1 and β^{2} = β + 1

Multiplying by

α^{10}

and

β^{10}

respectively:

α^{12} = α^{11} + α^{10} and β^{12} = β^{11} + β^{10}

Therefore:

a_{12} = p α^{12} + q β^{12} = p (α^{11} + α^{10}) + q (β^{11} + β^{10})

= (p α^{11} + q β^{11}) + (p α^{10} + q β^{10}) = a_{11} + a_{10}

JEE Advanced · 2017 · 21 May 2017 · Paper 2 · Single correct option