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

Hits: 14

 

Question

A sequence x1, x2, x3, …. ……. is given by:

x1=1,

xn+1=ax– 3, n≥1,

where a is a constant.

a.   Find an expression for x2 in terms of a.

b.   Show that x3=a2 – 3a – 3.

Given that x3=7,

c.   find the possible values of a.

Solution

a.
 

We are given the sequence as;

x1=1

xn+1= axn – 3

We are required to find x2.

xn+1= axn – 3

x1+1= ax1 – 3

x2=a(1) – 3

x2= a – 3

b.    

We are required to find x3.

xn+1= axn – 3

x2+1=ax2 – 3

x3=a(a – 3) – 3

x3= a2 -3a
-3

c.    

We have found in (b) that;

x3= a2 -3a -3

We are given that x3=7.

We are required to find p.

Substitute x3=7 in above equation.

7=a2 – 3a – 3

7-7= a2 – 3a -3 -7

a2 – 3a -10 = 0

a2 – 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.

Please follow and like us:
error0

Comments