A geometric series has first term a and common ratio r - Edexcel - A-Level Maths Pure - Question 9 - 2006 - Paper 2

To find the roots of the quadratic equation $2.5r^2 - 2.5r + 4 = 0$, we use the quadratic formula:

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

Here, $a = 2.5$, $b = -2.5$, and $c = 4$:

Calculating the discriminant:

$b^2 - 4ac = (-2.5)^2 - 4 \cdot 2.5 \cdot 4 = 6.25 - 40 = -33.75$

Since the discriminant is negative, there are no real values for r.

As the values of r are complex, we can use them to find a:

From i) $ar = 4$,

This implies:

$a = \frac{4}{r}$

With two complex values of r from earlier, calculate:

$a_1 = \frac{4}{r_1}, \; a_2 = \frac{4}{r_2}$.

We know the formula for the sum of the first n terms of a geometric series:

$S_n = a \frac{1 - r^n}{1 - r}$

Substituting for a:

$S_n = 25(1 - r) \frac{1 - r^n}{1 - r}$

This simplifies to:

$S_n = 25(1 - r^n)$

From here we can conclude:

$S_n = 25(1 - r).$

We set up the inequality:

$S_n > 24$

Substituting $S_n$ gives:

$25(1 - r) > 24$

Solving this, we have:

$1 - r > \frac{24}{25}$

Thus,

$r < \frac{1}{25}$

Finally, we plug in the values and use logs to isolate n:

$25(0.8)^n > 24$

$n > \frac{\log(0.04)}{\log(0.8)}$

Calculating gives:

$n \approx 15$.