A sequence $x_1,x_2,x_3,…$ 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 4 - 2012 - Paper 1

Now to find $x_3$, we substitute $n = 2$ into the recursive formula:

$x_3 = ax_2 + 5.$

Substituting our expression for $x_2$:

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

Expanding this gives:

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

Given that $x_5 = 41$, we first express $x_4$:

$x_4 = ax_3 + 5.$

Next, we use the expression for $x_3$:

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

Continuing this, we compute $x_5$:

$x_5 = ax_4 + 5 = a[a(a^2 + 5a + 5) + 5] + 5.$

Setting this equal to 41 gives the equation:

$a[a(a^2 + 5a + 5) + 5] + 5 = 41.$

Simplifying results in:

$a(a^3 + 5a^2 + 5a + 5) = 36,$

Now, let’s find $a$:

The equation $a(a^3 + 5a^2 + 5a + 5) = 36$ leads to finding roots.

This gives rise to possible solutions either by factoring or by substitution, leading to values of $a = 4$ and $a = -9$.