Series, Progressions and the Binomial Expansion: Question 10

Worked solution

Part (a): Passes sold on day 5

The daily sales form a geometric progression with first term $a=800$ and common ratio $r=\dfrac{1}{2}$ (each day’s sales are half the previous day’s). The $n$th term is

$u_n = ar^{n-1}.$

For day $5$ ($n=5$):

$u_5 = 800\times\left(\frac{1}{2}\right)^{4} = 800\times\frac{1}{16} = 50.$

Check: listing the daily sales directly, $800, 400, 200, 100, 50, \ldots$. The $5$th value in this list is indeed $50$, confirming the formula.

Part (b): Total passes sold over the 6 days

$S_n = \frac{a\bigl(1-r^n\bigr)}{1-r}$

$S_6 = \frac{800\bigl(1-(0.5)^6\bigr)}{1-0.5} = \frac{800(1-0.015625)}{0.5} = \frac{800\times0.984375}{0.5} = 800\times1.96875 = 1575.$

Check by adding the six daily figures directly: $800+400+200+100+50+25 = 1575$. This matches the sum formula exactly (note the $6$th day’s sales are $u_6=800\times(0.5)^5=25$).

Final answers

• (a) On day $5$, $\boxed{50}$ passes are sold.
• (b) The total number of passes sold over the $6$ days is $\boxed{1575}$.