Let p- q be integers and let - - be the roots of the equation- x2 - x - 1 - 0- where - For n - 0- 1- 2- - let an - p-n - q-n-FACT - If a and b are rational numbers and a - b-5 - 0- then a - 0 - b- a12 -

Let $p, q$ be integers and let $α, β$ be the roots of the equation $x^{2} - 2 x + 3 = 0$ where $α \neq β .$ For $n = 0, 1, 2, . . .,$ let $a_{n} = p α^{n} + q β^{n}$ and $a_{3} = - 10,$ then $p^{q} + q^{p} =$ ___

Let $p, q$ be integers and let $α, β$ be the roots of the equation $x^{2} - 2 x + 3 = 0$ where $α \neq β .$ For $n = 0, 1, 2, . . .,$ let $a_{n} = p α^{n} + q β^{n},$ then $a_{9} =$

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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+b√5=0, then a=0=b.a12=

The correct option is B a11+a10an+2−an+1−an=pαn(α2−α−1)+qβn(β2−β−1)=0So,a12−a11−a10=0a12=a11+a10

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 $n = 0, 1, 2, . . .$ let $a_{n} = p α^{n} + q β^{n}$ .
FACT : If $a$ and $b$ are rational numbers and $a + b \sqrt{5} = 0$ , then $a = 0 = b$ .
$a_{12} =$

The correct option is B $a_{11} + a_{10}$
$a_{n + 2} - a_{n + 1} - a_{n} = p α^{n} (α^{2} - α - 1) + q β^{n} (β^{2} - β - 1) = 0$
So,
$a_{12} - a_{11} - a_{10} = 0$
$a_{12} = a_{11} + a_{10}$