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 10 - 2016 - Paper 1

To find the amount John received on his 18th birthday, we need the 9th term of the sequence since his first gift was at age 10:

1. Using the formula for the n-th term:

   $t_9 = 60 + (9-1) imes 15 = 60 + 120 = 180$

Thus, John received £180 as a birthday gift on his 18th birthday.

To find the total gifts received up to the 21st birthday:

1. Calculate the 12th term:

   $t_{12} = 60 + (12 - 1) imes 15 = 225$

2. Now calculate the total for the first 12 terms using the sum formula:

   $S_n = \frac{n}{2} (t_1 + t_n)$

   There are 12 terms:

   $S_{12} = \frac{12}{2} (60 + 225) = 1710$

So the total gifts up to and including John's 21st birthday is £1710.

Given the total money received:

1. The sum of gifts can be expressed as:

   $S_n = \frac{n}{2} (60 + t_n)$

2. Setting this equal to £3375 gives:

   $\frac{n}{2} (60 + (60 + (n-1) \times 15)) = 3375$

   Simplifying leads to:

   $\frac{n}{2} (60 + 60 + 15n - 15) = 3375$

   or

   $\frac{n}{2} (105 + 15n) = 3375$

   Multiplying through by 2:

   $n(105 + 15n) = 6750$

   Reorganizing gives:

   $15n^2 + 105n - 6750 = 0$

   Dividing everything by 15 leads to:

   $n^2 + 7n - 450 = 0$

   which implies:

   $n^2 + 7n = 25 × 18$

To solve for n in the equation:

$n^2 + 7n - 450 = 0$

1. Use the quadratic formula:

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

   where $a = 1$, $b = 7$, and $c = -450$:

   $n = \frac{-7 \pm \sqrt{7^2 - 4 \times 1 \times (-450)}}{2 \times 1}$

   Simplifying the equation gives:

   $n = \frac{-7 \pm \sqrt{49 + 1800}}{2} = \frac{-7 \pm \sqrt{1849}}{2}$

   Finding the square root:

   $\sqrt{1849} = 43$

   Thus:

   $n = \frac{-7 + 43}{2} = 18$ (the positive root only)

2. Since John received his first gift at age 10, John's age when he received £3375 is:

   $10 + n = 10 + 18 = 28$

Thus, John was 28 years old at that time.