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

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

A sequence $x_1,x_2,x_3,...$ is defined by $x_1 = 1$ $x_{n+1} = a x_n + 5,$ $ n > 1$ where $a$ is a constant. (a) Write down an expression for $x_2$ in terms of... show full transcript

Worked Solution & Example Answer:A sequence $x_1,x_2,x_3,...$ is defined by $x_1 = 1$ $x_{n+1} = a x_n + 5,$ $ n > 1$ where $a$ is a constant - Edexcel - A-Level Maths Pure - Question 5 - 2012 - Paper 1

Step 1

Write down an expression for $x_2$ in terms of $a$.

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Answer

Using the definition of the sequence, we have:

x2=ax1+5x_2 = a x_1 + 5

Substituting x1=1x_1 = 1 gives:

x2=a(1)+5=a+5x_2 = a(1) + 5 = a + 5

Step 2

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

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Answer

To find x3x_3, we use the expression we found for x2x_2:

x3=ax2+5x_3 = a x_2 + 5

Substituting x2=a+5x_2 = a + 5:

x3=a(a+5)+5x_3 = a(a + 5) + 5

Expanding this gives:

x3=a2+5a+5x_3 = a^2 + 5a + 5

Step 3

Given that $x_5 = 41$, find the possible values of $a$.

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Answer

From earlier, we have:

x5=ax4+5x_5 = a x_4 + 5

Continuing this pattern, we find:

  1. x4=ax3+5x_4 = a x_3 + 5; substituting x3=a2+5a+5x_3 = a^2 + 5a + 5 gives:

    x4=a(a2+5a+5)+5x_4 = a(a^2 + 5a + 5) + 5

    Expanding will yield:

    x4=a3+5a2+5a+5x_4 = a^3 + 5a^2 + 5a + 5

  2. Then substitute x4x_4 into x5x_5:

    x5=a(a3+5a2+5a+5)+5x_5 = a(a^3 + 5a^2 + 5a + 5) + 5

Setting this equal to 41 leads to:

a4+5a3+5a2+5a+5=41a^4 + 5a^3 + 5a^2 + 5a + 5 = 41

Subtracting 41 gives:

a4+5a3+5a2+5a36=0a^4 + 5a^3 + 5a^2 + 5a - 36 = 0

This is a polynomial equation in terms of aa. Solving this equation using numerical methods or factoring will yield the possible values of aa. The roots can be checked for suitable values.

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