The adult population of a town is 25 000 at the end of Year 1 - Edexcel - A-Level Maths Pure - Question 10 - 2010 - Paper 3

The common ratio $r$ of the geometric sequence is given by the annual increase. Here, since the population increases by 3% each year:

$r = 1 + 0.03 = 1.03.$

We start with the population model:

$P_N = P_0 \times (1.03)^{N-1}$

Setting $P_N > 40000$ gives us:

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

By dividing both sides by 25000:

$(1.03)^{N-1} > \frac{40000}{25000} = 1.6.$

Now, applying log on both sides:

$\log((1.03)^{N-1}) > \log(1.6).$ Using the property of logarithms:

$(N-1) \log(1.03) > \log(1.6),$ which simplifies to:

$(N-1) \log_{1.03} > \log_{1.6}.$

From the previous step, we have:

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

Now we can isolate $N$:

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

Calculating those logs using a calculator:

$\log(1.6) \approx 0.2041, \ \log(1.03) \approx 0.0128.$

Calculating:

$N - 1 > \frac{0.2041}{0.0128} \approx 15.91,$

Thus, since $N - 1$ must be an integer:

$N > 16.91 \implies N = 17.$

To find the total amount contributed to the charity fund:

Each year, each member of the population contributes £1. We need to calculate the population for every year from Year 1 to Year 10.

The adult population at Year N is given by:

$P_N = 25000 \times (1.03)^{N-1}$

We can calculate the total amount contributed:

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

This is a geometric series with:

• First term $a = 25000$
• Common ratio $r = 1.03$
• Number of terms $n = 10$.

The sum of a geometric series can be calculated using the formula:

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

Thus,

$Total = 25000 \frac{1.03^{10} - 1}{1.03 - 1}.$

Calculating:

First, find $1.03^{10} \approx 1.3439$. Then:

$Total \approx 25000 \frac{1.3439 - 1}{0.03} \approx 25000 \times 11.4633 \approx 286582.65.$

Rounding to the nearest £1 000 gives £287 000.