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

Abbie's total contributions after n years are given by the sum of an arithmetic series:

$S_n = \frac{n}{2} \times (2a + (n - 1)d)$

Where:

• ( a = 500 )
• ( d = 200 )
• This total is equal to £67200.

Substituting the values into the formula gives:

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

This simplifies to: $S_n = \frac{n}{2} \times (1000 + 200n - 200) = 67200$

$S_n = \frac{n}{2} \times (800 + 200n) = 67200$

Multiplying through by 2: $n(800 + 200n) = 134400$

Expanding this: $800n + 200n^2 = 134400$

Rearranging gives: $200n^2 + 800n - 134400 = 0$

Dividing through by 200 simplifies to: $n^2 + 4n - 672 = 0$

Now, relating this to the requirement:

$24 \times 28 = 672$

We have thus shown that: $n^2 + 4n - 24 \times 28 = 0$.

To solve the quadratic equation:

$n^2 + 4n - 672 = 0$

We can apply the quadratic formula:

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

Where:

• ( a = 1 )
• ( b = 4 )
• ( c = -672 )

Calculating the discriminant:

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

Now substituting into the quadratic formula:

$n = \frac{-4 \pm \sqrt{2704}}{2}$

Calculating ( \sqrt{2704} ):

$\sqrt{2704} = 52$

Thus, we have:

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

Calculating these gives us:

1. ( n = \frac{48}{2} = 24 )
2. ( n = \frac{-56}{2} = -28 ) (discarded as invalid)

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