A sequence $x_1, x_2, x_3, \, \ldots$ is defined by $x_1 = 1$ $x_{n+1} = ax_n + 5$, $n > 1$ where $a$ is a constant - Edexcel - A-Level Maths Pure - Question 6 - 2012 - Paper 1

To find $x_3$, we use the definition:

$x_3 = ax_2 + 5.$
Substituting for $x_2$, we have:

$x_3 = a(a + 5) + 5 = a^2 + 5a + 5.$

We know that:

$x_4 = ax_3 + 5.$
Substituting for $x_3$ gives:

$x_4 = a(a^2 + 5a + 5) + 5.$
Then,

$x_5 = ax_4 + 5 = a[a(a^2 + 5a + 5) + 5] + 5.$
Given that $x_5 = 41$ and by simplifying:

Opening up the equation, we have:

$41 = a(a^3 + 5a^2 + 5a + 5) + 5.$

Rearranging, we arrive at:

$a(a^3 + 5a^2 + 5a) = 41 - 5 = 36.$

This leads us to find a suitable polynomial to solve for $a$. We can try:

$a^4 + 5a^3 + 5a^2 - 36 = 0.$

Using methods like synthetic division or the quadratic formula can yield potential roots for $a$, leading to the possible values of $a$. After solving, we ascertain the values of $a$ as:

$a = 4 \text{ or } a = -9.$