On John’s 10th birthday he received the first of an annual birthday gift of money from his uncle - Edexcel - A-Level Maths Pure - Question 11 - 2016 - Paper 1

The amount of the gift on any birthday can be expressed as: $t_n = a + (n - 1)d$ For John's 18th birthday ($n = 8$ as starting from age 10): $t_8 = 60 + (8 - 1)15 = 60 + 105 = £165$

Therefore, John received £165 on his 18th birthday.

To find the total gifts received by his 21st birthday, we consider:

• Number of terms, $n = 12$ (ages 10 to 21 inclusive).

Using the sum formula again: $S_n = \frac{n}{2} \times (2a + (n-1)d)$ Calculating: $S_{12} = \frac{12}{2} \times (2 \times 60 + (12-1) \times 15)$ $= 6 \times (120 + 165) = 6 \times 285 = 1710$

Thus, total gifts up to John's 21st birthday is £1710.

Using the total amount John received, £3375, We apply the sum formula: $S_n = \frac{n}{2} \times (2a + (n-1)d) = £3375$ Substituting values $a=60$, $d=15$: $\frac{n}{2} \times (2 \times 60 + (n-1) \times 15) = 3375$ To simplify, we obtain: $3375 = \frac{n}{2} \times (120 + 15n - 15)$ $= \frac{n}{2} \times (15n + 105)$ $6750 = n(15n + 105)$ Rearranging gives: $15n^2 + 105n - 6750 = 0$ Dividing through by 15 yields: $n² + 7n = 450$ Thus, $450 = 25 \times 18$.

To solve $n² + 7n = 450$, we rearrange to: $n^2 + 7n - 450 = 0$ Using the quadratic formula, $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1, b=7, c=-450$: $n = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-450)}}{2 \times 1}$ $= \frac{-7 \pm \sqrt{49 + 1800}}{2}$ $= \frac{-7 \pm \sqrt{1849}}{2}$ $= \frac{-7 \pm 43}{2}$ This results in two possible solutions: $n = 18$ or $n = -25$. Since $n$ must be positive, $n = 18$. Thus, John received 18 gifts, and since he started receiving gifts at age 10, he is now: $10 + 18 = 28$ John is 28 years old at this time.