A geometric series has first term a and common ratio r - Edexcel - A-Level Maths Pure - Question 10 - 2006 - Paper 2

To find the roots of the equation:

$25r^2 - 25r + 4 = 0$

Using the quadratic formula, where $a = 25$, $b = -25$, and $c = 4$:

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

Calculating the discriminant:

$b^2 - 4ac = (-25)^2 - 4 \times 25 \times 4 = 625 - 400 = 225$

Now substituting back:

$r = \frac{25 \pm \sqrt{225}}{50} = \frac{25 \pm 15}{50}$

Thus, the two possible values are:

$r = 0.8 \quad \text{and} \quad r = 0.2$

With the values of r found, we substitute back to find a:

1. For $r = 0.8$:

   $a = \frac{4}{0.8} = 5$

2. For $r = 0.2$:

   $a = \frac{4}{0.2} = 20$

Thus, the corresponding two values of a are 5 and 20.

The sum of the first n terms of a geometric series is:

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

Replacing a with $25(1 - r)$ (from the earlier relation):

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

This confirms our formula for $S_n$.

Using the formula:

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

We need:

$25(1 - r^{n-1}) > 24$

This simplifies to:

$1 - r^{n-1} > 0.96$ \therefore\ r^{n-1} < 0.04$$

Now substituting the larger value of r, $0.8$:

$0.8^{n-1} < 0.04$

Taking logarithm:

$\log(0.8^{n-1}) < \log(0.04)$

This gives:

$(n-1) \log(0.8) < \log(0.04)$

Calculating:

$\log(0.04) \approx -1.39794\ ext{and} \ \log(0.8) \approx -0.09691$

Substituting these values:

$n - 1 > \frac{1.39794}{0.09691} \approx 14.43\ herefore\ n > 15.43$

Thus, the smallest integer value of n is 15.