A trading company made a profit of £50 000 in 2006 (Year 1) - Edexcel - A-Level Maths Pure - Question 9 - 2007 - Paper 2

To show that ( P_n > 200000 ), we start with:

$50000 r^{n-1} > 200000$

Dividing both sides by 50000 gives:

$r^{n-1} > 4$

Taking the logarithm of both sides:

$\log(r^{n-1}) > \log(4)$

This simplifies to:

$(n-1) \log(r) > \log(4)$

Dividing by (\log(r)) (which is positive since r > 1):

$n - 1 > \frac{\log(4)}{\log(r)}$

Thus:

$n > \frac{\log(4)}{\log(r)} + 1$

Using the model with r = 1.09, we substitute into our inequality:

$n > \frac{\log(4)}{\log(1.09)} + 1$

Calculating the right-hand side:

1. Calculate ( \log(4) ) and ( \log(1.09) ):

   ( \log(4) \approx 0.6021 ), ( \log(1.09) \approx 0.0374 )

2. Substitute these values:

   $n > \frac{0.6021}{0.0374} + 1 \approx 17.0$

Thus, the profit will first exceed £200,000 in Year 18 (2024).

The total profit over 10 years can be found using the formula for the sum of a geometric series:

$S_n = a \frac{1 - r^n}{1 - r}$

where:

• a is the first term (in this case, £50,000),
• r is the common ratio (1.09),
• n is the number of terms (10).

Substituting the values:

$S_{10} = 50000 \frac{1 - 1.09^{10}}{1 - 1.09}$

Calculating ( 1.09^{10} \approx 2.3674 ):

$S_{10} = 50000 \frac{1 - 2.3674}{-0.09} \approx 50000 \times 15.7 \approx 785000 \approx £760000$

Thus, the total profit from 2006 to 2015 is approximately £760,000.