The adult population of a town is 25 000 at the end of Year 1 - Edexcel - A-Level Maths Pure - Question 1 - 2010 - Paper 3

The common ratio, $r$, in a geometric sequence, where the population increases by 3% each year, is:

$r = 1 + 0.03 = 1.03$

Starting with the population model, we have:

$\frac{P_N}{P_1} = (1 + r)^{N - 1}$

Given that $P_1 = 25000$ and $P_N > 40000$, we set up the inequality:

$\frac{40000}{25000} > (1.03)^{N - 1}$

This simplifies to:

$1.6 > (1.03)^{N - 1}$

Taking the logarithm of both sides yields:

$\log(1.6) > (N - 1) \log(1.03)$

Rearranging gives:

$(N - 1) \log(1.03) > \log(1.6)$

To find $N$, we first rewrite the inequality found in part (c):

$(N - 1) > \frac{\log(1.6)}{\log(1.03)}$

Calculating $\log(1.6)$ and $\log(1.03)$, we find:

• $\log(1.6) \approx 0.2041$
• $\log(1.03) \approx 0.01283$

Thus, substituting into the equation gives:

$(N - 1) > \frac{0.2041}{0.01283} \approx 15.9$

This implies:

$N > 16.9$

Since $N$ must be a whole number, $N = 17$.

The total amount given to the charity fund is calculated using the sum of the populations from Year 1 to Year 10:

Let $P_n$ represent the population at the end of Year n. Each year, the population is:

$P_n = 25000 \times (1.03)^{n-1}$

The total over 10 years can be computed as:

$\text{Total} = \sum_{n=1}^{10} P_n = 25000 \times \sum_{n=0}^{9} (1.03)^n$.

The sum of a geometric series can be calculated as:

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

Using $a = 1$, $r = 1.03$, and $n = 10$:

$\sum = \frac{1 - (1.03)^{10}}{1 - 1.03} = \frac{1 - 1.3439}{-0.03} \approx 11.464$

So,

$\text{Total Amount} = 25000 \times 11.464 \approx 286,600$

Rounding to the nearest £1,000, the final answer is approximately £287,000.