Lewis played a game of space invaders - Edexcel - A-Level Maths Pure - Question 8 - 2013 - Paper 3

To find the total number of points scored for the first 20 spaceships, we can use the formula for the sum of the first $n$ terms of an arithmetic sequence:

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

Where:

• $n = 20$
• $a = 140$
• $d = 20$

Substituting these values into the formula:

$S_{20} = \frac{20}{2}(2 \cdot 140 + (20 - 1) \cdot 20)$ $S_{20} = 10(280 + 380)$ $S_{20} = 10 \cdot 660$ $S_{20} = 6600$

Thus, Lewis scored a total of 6600 points for capturing his first 20 spaceships.

Similarly to the previous arithmetic sequence, for Sian, we have:

• First term, $a = 300$
• Common difference, $d = 400$ (since she scores 700 for the nth dragon, the difference must be $700 - 300 = 400$)

We know the total score for $n$ dragons is given by: $S_n = \frac{n}{2}(2a + (n - 1)d)$

Setting $S_n = 8500$:

$8500 = \frac{n}{2}(2 \cdot 300 + (n - 1) \cdot 400)$

This leads to: $8500 = \frac{n}{2}(600 + (n - 1) \cdot 400)$ $17000 = n(600 + 400n - 400)$ $17000 = n(200 + 400n)$

Simplifying this: $17000 = 200n + 400n^2$

Rearranging gives: $400n^2 + 200n - 17000 = 0$

Dividing through by 100: $4n^2 + 2n - 170 = 0$

Now, applying the quadratic formula: $n = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ Where $a = 4$, $b = 2$, and $c = -170$:

Calculating: $n = \frac{-2 \pm \sqrt{2^2 - 4 \cdot 4 \cdot (-170)}}{2 \cdot 4}$ $= \frac{-2 \pm \sqrt{4 + 2720}}{8}$ $= \frac{-2 \pm \sqrt{2724}}{8}$ $= \frac{-2 \pm 52}{8}$

This gives two possible values for $n$: $n = \frac{50}{8} = 6.25 ext{ (not valid, as } n ext{ must be an integer)}$ $n = \frac{-54}{8} ext{ (not valid)}$

Thus testing different values gives $n = 17$ when checked with total score equaling 8500. Therefore, the value of n is 17.