1. a12 = ....... 2. If a4 = 28, then p + 2q = .....

Question

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 a_n = pαⁿ + qβⁿ .

FACT: If a and b are rational numbers and a + b√5 = 0, then a = 0 = b.

1.  a₁₂  = .......

(A)  a₁₁ - a₁₀ 

(B)  a₁₁ + a₁₀ 

(C)  2a₁₁ + a₁₀

(D)  a₁₁ + 2a₁₀

2.  If a₄ = 28, then p + 2q = .....

(A)  21 

(B)  14 

(C)  7 

(D)  12

Answer 1

Correct option 1. (B) a₁₁ + a₁₀  2. (D) 12

1.  It is given that x² - x - 1 = 0. 

Also, α and β are roots of equation and α≠β.

Let p and q be integers and pαⁿ + qβⁿ = a_n.

Since α and β are the roots of x² = x + 1, we get

α₁₁ + α₁₀ = pα¹¹ + qβ¹¹ + pα¹⁰ + qβ¹⁰

⁼  pα¹¹ +  pα¹⁰ + qβ¹¹ + qβ¹⁰

= pα¹⁰(α + 1) + qβ¹⁰(β + 1)

= pα¹² + qβ¹⁰β²

⁼ pα¹² + qβ¹² = q₁₂

That is,

a₁₁ + a₁₀ = a₁₂. 

2 . It is given that a₄ = 28. Using a_n = pαⁿ + qβⁿ, we get

a_n - a_{n - 1} 

Therefore, 

Now, from x² - x - 1 = 0, the roots of the equation are

It is given that if a and b are rational numbers and a b + = 5 0; then, a = 0 = 6. Therefore,