The sum to infinity of a geometric series is 96 - AQA - A-Level Maths Pure - Question 8 - 2020 - Paper 3

To solve this problem, we begin with the formula for the sum to infinity of a geometric series:

the sum $S$ is given by:
$S = \frac{a}{1 - r}$
Where $a$ is the first term and $r$ is the common ratio.
Given that the sum to infinity is 96, we can write:
$\frac{a}{1 - r} = 96$

We also know that the second term is 18, which can be expressed as:
$u_2 = ar = 18$

Now we have two equations:

1. $\frac{a}{1 - r} = 96$
2. $ar = 18$

From the second equation, we can express $a$ in terms of $r$:
$a = \frac{18}{r}$

Substituting this into the first equation gives:
$\frac{\frac{18}{r}}{1 - r} = 96$

Multiplying both sides by $r(1 - r)$ leads to:
$18 = 96r(1 - r)$
Expanding this gives:
$18 = 96r - 96r^2$
Rearranging leads us to:
$96r^2 - 96r + 18 = 0$

We can solve this quadratic equation using the quadratic formula:
$r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$
Where $a = 96$, $b = -96$, and $c = 18$.
Substituting in these values:
$r = \frac{96 \pm \sqrt{(-96)^2 - 4 \cdot 96 \cdot 18}}{2 \cdot 96}$
This calculation leads to two potential solutions for $r$. After substituting back, we find two possible values for $a$:
a) $a = 24$ and $r = \frac{3}{4}$;
b) $a = 72$ and $r = \frac{1}{4}$.

Given that the first term must be less than 30, we conclude:
$a = 24, \quad r = \frac{3}{4}$.