# Past Papers’ Solutions | Edexcel | AS & A level | Mathematics | Core Mathematics 1 (C1-6663/01) | Year 2008 | June | Q#5

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**Question**

A sequence x_{1}, x_{2}, x_{3}, …. ……. is given by:

x_{1}=1,

x_{n+1}=ax_{n }– 3, n≥1,

where a is a constant.

**a. **Find an expression for x_{2} in terms of a.

**b. **Show that x_{3}=a^{2} – 3a – 3.

Given that x_{3}=7,

**c. **find the possible values of a.

**Solution**

**a.
**

We are given the sequence as;

x_{1}=1

x_{n+1}= ax_{n} – 3

We are required to find x_{2}.

x_{n+1}= ax_{n} – 3

x_{1+1}= ax_{1} – 3

x_{2}=a(1) – 3

x_{2}= a – 3

**b. **

We are required to find x_{3}.

x_{n+1}= ax_{n} – 3

x_{2+1}=ax_{2} – 3

x_{3}=a(a – 3) – 3

x_{3}= a^{2} -3a

-3

**c. **

We have found in (b) that;

x_{3}= a^{2} -3a -3

We are given that x_{3}=7.

We are required to find p.

Substitute x_{3}=7 in above equation.

7=a^{2} – 3a – 3

7-7= a^{2} – 3a -3 -7

a^{2} – 3a -10 = 0

a^{2} – 5a +2a -10 = 0

a(a – 5) + 2(a – 5) = 0

(a + 2)(a – 5) = 0

Now we have two options.

a + 2=0 |
a – 5=0 |

a + 2 -2=0-2 |
a – 5 + 5=5 |

a=-2 |
a=5 |

We are given that a≥1, therefore,a=5.

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