Each year, Abbie pays into a savings scheme - Edexcel - A-Level Maths Pure - Question 9 - 2013 - Paper 2

Abbie's total payments over n years is given by the formula for the sum of an arithmetic series: $S_n = \frac{n}{2} \times (2a + (n-1)d)$ Substituting the values: $S_n = \frac{n}{2} \times (2 \times 500 + (n-1) \times 200)$ $= \frac{n}{2} \times (1000 + 200n - 200)$ $= \frac{n}{2} \times (200n + 800)$ $= 100n^2 + 400n$

Setting this equal to £67200: $100n^2 + 400n = 67200$ Dividing by 100: $n^2 + 4n = 672$ Rearranging gives: $n^2 + 4n - 672 = 0$ Now, recognizing that 672 = 24 × 28 yields: $n^2 + 4n - 24 \times 28 = 0$

To solve the quadratic equation $n^2 + 4n - 672 = 0$, we can use the quadratic formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Here:

• $a = 1$
• $b = 4$
• $c = -672$

Calculating the discriminant: $b^2 - 4ac = 4^2 - 4 \times 1 \times (-672) = 16 + 2688 = 2704$

Now, applying the values into the quadratic formula: $n = \frac{-4 \pm \sqrt{2704}}{2}$ $= \frac{-4 \pm 52}{2}$ Considering the positive solution: $n = \frac{48}{2} = 24$ Thus, Abbie pays into the savings scheme for 24 years.