In an arithmetic series, the first term is a and the common difference is d - Edexcel - A-Level Maths Pure - Question 15 - 2022 - Paper 1

Given the savings:

• The first term, a = 10,
• The second term can be calculated: 9.20 = 10 - 0.8,
• The common difference, d = -0.8.

To express the total savings after n weeks:

1. The total savings can be calculated using the sum formula: $S_n = \frac{n}{2}[2a + (n-1)d]$
2. Substitute the values: $S_n = \frac{n}{2}[2(10) + (n-1)(-0.8)]$
3. Setting this equal to 64 (the cost of the printer): $\frac{n}{2}[20 - 0.8(n-1)] = 64$
4. Clearing the fractions and simplifying leads to: $n^2 - 26n + 160 = 0$.

The quadratic equation is: $n^2 - 26n + 160 = 0$ To solve for n, use the quadratic formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where: a = 1, b = -26, c = 160.

1. Calculate the discriminant: $b^2 - 4ac = (-26)^2 - 4(1)(160) = 676 - 640 = 36$
2. Calculate the roots: $n = \frac{26 \pm 6}{2}$ gives: $n = 16 \; \text{or} \; n = 10.$

James takes n = 10 weeks to save exactly £64. As the calculation shows both 10 and 16 as potential solutions, but since he begins saving from week 1, 10 weeks is the minimal solution that allows him to reach the total amount needed for the printer.