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JEE Advance - Mathematics (2017 - Paper 2 Offline - No. 18)

Let p, q be integers and let $$\alpha $$, $$\beta $$ be the roots of the equation, x2 $$-$$ x $$-$$ 1 = 0 where $$\alpha $$ $$ \ne $$ $$\beta $$. For n = 0, 1, 2, ........, let an = p$$\alpha $$n + q$$\beta $$n.

FACT : If a and b are rational numbers and a + b$$\sqrt 5 $$ = 0, then a = 0 = b.
Let p, q be integers and let $$\alpha $$, $$\beta $$ be the roots of the equation, x2 $$-$$ x $$-$$ 1 = 0 where $$\alpha $$ $$ \ne $$ $$\beta $$. For n = 0, 1, 2, ........, let an = p$$\alpha $$n + q$$\beta $$n.

FACT : If a and b are rational numbers and a + b$$\sqrt 5 $$ = 0, then a = 0 = b.
Let p, q be integers and let $$\alpha $$, $$\beta $$ be the roots of the equation, x2 $$-$$ x $$-$$ 1 = 0 where $$\alpha $$ $$ \ne $$ $$\beta $$. For n = 0, 1, 2, ........, let an = 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 a4 = 28, then p + 2q =
14
7
21
12

Explanation

$$\alpha = {{1 + \sqrt 5 } \over 2}$$,

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

$${a_4} = {a_3} + {a_2}$$

$$ = 2{a_2} + {a_1}$$

$$ = 3{a_1} + 2{a_0}$$


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

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

$$ \therefore $$ p $$-$$ q = 0

and $$(p + q) \times {7 \over 2} = 28$$

$$ \Rightarrow $$ p + q = 8

$$ \Rightarrow $$ p = q = 4

$$ \therefore $$ p + 2q = 12

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