A sequence $x_1, x_2, x_3, \, \ldots$ is defined by $x_1 = 1,$ where $a$ is a constant. $x_{n+1} = ax_n - 3, \, n > 1.$ (a) Find an expression for $x_n$ in terms of $a$. (b) Show that $x_3 = a^2 - 3a - 3.$ (c) Given that $x_3 = 7,$ find the possible values of $a.$

A-Level Edexcel Maths Pure Arithmetic Sequences & Series: A sequence $x_1, x_2, x

A sequence $x_1, x_2, x_3, \, \ldots$ is defined by $x_1 = 1,$ where $a$ is a constant - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 1

Question 7

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A sequence $x_1, x_2, x_3, \, \ldots$ is defined by $x_1 = 1,$ where $a$ is a constant. $x_{n+1} = ax_n - 3, \, n > 1.$ (a) Find an expression for $x_n$ in terms o... show full transcript

Worked Solution & Example Answer:A sequence $x_1, x_2, x_3, \, \ldots$ is defined by $x_1 = 1,$ where $a$ is a constant - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 1

Step 1

Find an expression for $x_n$ in terms of $a$.

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Answer

To find an expression for $x_n$ , we can compute the first few terms of the sequence:

For $n=1$ :

$x_1 = 1$
For $n=2$ :

$x_2 = ax_1 - 3 = a(1) - 3 = a - 3$
For $n=3$ :

$x_3 = ax_2 - 3 = a(a - 3) - 3 = a^2 - 3a - 3$

Continuing this pattern leads to the conclusion that the expression takes the form of a recursion. Therefore, we can generalize it for any $n$ :

$x_n = ax_{n-1} - 3$

Step 2

Show that $x_3 = a^2 - 3a - 3.$

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Answer

From the calculations in part (a), we already determined that:

$x_3 = ax_2 - 3$ Substituting for $x_2$ , we have:

$x_3 = a(a - 3) - 3 = a^2 - 3a - 3$

Thus, we have shown that $x_3 = a^2 - 3a - 3$ .

Step 3

Given that $x_3 = 7,$ find the possible values of $a.$

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Answer

We set the equation derived from part (b) equal to $7$ :

$a^2 - 3a - 3 = 7$

This simplifies to:

$a^2 - 3a - 10 = 0$

To solve this quadratic equation, we can factor or use the quadratic formula:

$a = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(-10)}}{2(1)}$

$a = \frac{3 \pm \sqrt{9 + 40}}{2} = \frac{3 \pm \sqrt{49}}{2} = \frac{3 \pm 7}{2}$

So the solutions are:

$a = \frac{10}{2} = 5 \quad \text{and} \quad a = \frac{-4}{2} = -2$

Thus, the possible values of $a$ are $5$ and $-2$ .

A sequence $x_1, x_2, x_3, \, \ldots$ is defined by $x_1 = 1,$ where $a$ is a constant - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 1

Worked Solution & Example Answer:A sequence $x_1, x_2, x_3, \, \ldots$ is defined by $x_1 = 1,$ where $a$ is a constant - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 1

Find an expression for $x_n$ in terms of $a$.

Show that $x_3 = a^2 - 3a - 3.$

Given that $x_3 = 7,$ find the possible values of $a.$

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