Year in a sequence (1)

We have a linear recurrence here with $a_1 = 11$ and $a_2 = a_1 + 9 = 11 + 9 = 20.$ Taking,

$a_n = a_{n-1} + 9;$

$a_{n+1} = a_n + 9$

in a combination produces $a_{n+1} = a_n + (a_n - a_{n-1}) \Rightarrow a_{n+1} - 2a_n + a_{n-1} = 0$ with a characteristic equation of $r^2 - 2r + 1 = 0 \Rightarrow (r-1)^2 = 0 \Rightarrow r = 1$ . So now our general solution becomes $a_n = An + B$ , $(A, B \in \mathbb{R})$ .

We can now solve for the constants using our initial conditions $a_1 = 11, a_2 = 20:$

$11 = A + B$ and $20 = 2A + B$

which yields $A = 9, B = 2$ , or $a_n = 9n + 2$ . Hence, $a_{2017} = 9(2017) + 2 = \boxed{18,155}.$

Each term exceeds the previous one by $9$ , then $a_{2017}= 9\times 2016+11 = 18155$ .

I multiplied by $2016$ because $a_1= 11$ .