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 $a=0=b$. If $a_{4}=28$, then $p+2q=$

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

.

If

a_{4} = 28

, then

p + 2 q =

Answer: (D)

From the recurrence relation

a_{n} = a_{n - 1} + a_{n - 2}

, we can express:

a_{4} = 2 a_{0} + 3 a_{1}

Now,

a_{0} = p + q

and

a_{1} = p α + q β

.

Since

α + β = 1

(sum of roots) and

α β = - 1

(product of roots):

a_{1} = p α + q β

Using the recurrence:

a_{2} = a_{1} + a_{0} = (p α + q β) + (p + q)

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

a_{4} = a_{3} + a_{2} = 3 a_{1} + 2 a_{0}

Given

a_{4} = 28

3 (p α + q β) + 2 (p + q) = 28

Since

α = \frac{1 + 5}{2}

and

β = \frac{1 - 5}{2}

3 p α + 3 q β + 2 p + 2 q = 28

(3 α + 2) p + (3 β + 2) q = 28

For this to hold for integer

p, q

, and using the given FACT, we find

p = q = 4

.

Therefore,

p + 2 q = 4 + 8 = 12

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