JEE Advanced 2017

Given that $x^2 - x - 1 = 0$ $\implies x^2 = x + 1$ $\implies x^n = x^{n-1} + x^{n-2}$ , where $n\ge 2$ is a positive integer. Since $\alpha$ and $\beta$ are roots of $x^2 - x - 1 = 0$ , it means that $\alpha^n = \alpha^{n-1}+\alpha^{n-2}$ and $\beta^n = \beta^{n-1}+\beta^{n-2}$ . Then we have:

$\begin{aligned} a_n & = p\alpha^n + q \beta^n \\ & = p \left(\alpha^{n-1}+\alpha^{n-2}\right) + q \left(\beta^{n-1}+\beta^{n-2}\right) \\ & = p \alpha^{n-1} + q \beta^{n-1} + p \alpha^{n-2} + q \beta^{n-2} \\ & = a_{n-1} + a_{n-2} & \small \color{#3D99F6} \text{For }n = 12 \\ \implies a_{12} & = \boxed{a_{11} + a_{10}} \end{aligned}$