Each year, Abbie pays into a savings scheme - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 2

Abbie pays into the scheme for n years. The total amount paid can be expressed as:

$$ext{Total} = \frac{n}{2} \times (\text{first payment} + \text{last payment})$$

The last payment, made in year n, is:

$$500 + 200(n-1)$$

Thus, using the first payment as £500, the expression becomes:

$$\frac{n}{2} \times (500 + (500 + 200(n-1))) = 67200$$

This simplifies to:

$$\frac{n}{2} \times (1000 + 200(n-1)) = 67200$$ $$\frac{n}{2} \times (1000 + 200n - 200) = 67200$$ $$\frac{n}{2} \times (800 + 200n) = 67200$$ $$n(800 + 200n) = 134400$$

Rearranging gives:

$$200n^2 + 800n - 134400 = 0$$

Dividing through by 8:

$$25n^2 + 100n - 16800 = 0 ext{, which simplifies to } n^2 + 4n - 24 imes 28 = 0$$

To solve the quadratic equation $n^2 + 4n - 24 \times 28 = 0$, we can apply the quadratic formula:

$$n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

Here, a = 1, b = 4, and c = -672. Substituting these values gives:

$$n = \frac{-4 \pm \sqrt{4^2 - 4 \cdot 1 \cdot (-672)}}{2 imes 1}$$

Calculating the discriminant:

$$4^2 - 4 \cdot 1 \cdot -672 = 16 + 2688 = 2704$$

Now, substituting back into the formula:

$$n = \frac{-4 \pm \sqrt{2704}}{2} ext{, where } \sqrt{2704} = 52$$

Thus:

$$n = \frac{-4 + 52}{2} ext{ or } \frac{-4 - 52}{2}$$

Calculating both cases gives:

1. $n = \frac{48}{2} = 24$
2. $n = \frac{-56}{2}, which is not valid$

Therefore, Abbie pays into the savings scheme for 24 years.