A car was purchased for £18 000 on 1st January - Edexcel - A-Level Maths Pure - Question 8 - 2010 - Paper 4

The car's value falls below £1000. We use the same formula:

$V = 18000 imes (0.8)^n$

Setting $V < 1000$:

$18000 imes (0.8)^n < 1000$

Dividing both sides by 18000 gives:

(0.8)^n < rac{1000}{18000}

Calculating the right side:

(0.8)^n < rac{1}{18}

Taking the logarithm of both sides:

n imes ext{log}(0.8) < ext{log}rac{1}{18}

Solving for $n$:

n ext{ is approximately } 13 ext{ years.} $$

The cost of the maintenance scheme follows a geometric progression with a first term of £200 and a common ratio (increase by 12%) of 1.12. The formula for the nth term is:

$a_n = a imes r^{(n-1)}$

where:

• $a = 200$
• $r = 1.12$
• $n = 5$

Calculating:

$a_5 = 200 imes (1.12)^{4}$

which approximately equals £314.70.

The total cost can be calculated as:

S_n = a imes rac{1 - r^n}{1 - r}

Applying:

• $a = 200$
• $r = 1.12$
• $n = 15$

Calculating:

S_{15} = 200 imes rac{1 - (1.12)^{15}}{1 - 1.12}

This results in approximately £977.42. Therefore, the total cost is £977.42.