The first three terms of a geometric series are 4p, (3p + 15) and (5p + 20) respectively, where p is a positive constant - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 5

To solve the quadratic equation $11p^2 - 10p - 225 = 0$, we can use the quadratic formula:

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

where $a = 11$, $b = -10$, and $c = -225$.

Calculating the discriminant:

$b^2 - 4ac = (-10)^2 - 4(11)(-225) = 100 + 9900 = 10000$

Thus:

$p = \frac{10 \pm 100}{22}$

Evaluating the possible solutions:

1. $p = \frac{110}{22} = 5$
2. $p = \frac{-90}{22} = -4.09$

Since p is a positive constant, we have:

$p = 5$

The common ratio $r$ can be found using the first two terms:

From the first term 4p:

$r = \frac{3p + 15}{4p}$

Substituting $p = 5$:

$r = \frac{3(5) + 15}{4(5)} = \frac{15 + 15}{20} = \frac{30}{20} = \frac{3}{2}$

Thus, the common ratio of the series is:

$r = \frac{3}{2}$

The sum $S_n$ of the first n terms of a geometric series is given by:

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

Where:

• $a$ is the first term (which is $4p = 20$ if $p=5$)
• $r$ is the common ratio ($\frac{3}{2}$)
• $n$ is the number of terms (10)

Substituting these values into the formula gives:

$S_{10} = 20 \frac{1 - (\frac{3}{2})^{10}}{1 - \frac{3}{2}}$

Calculating:

$S_{10} = 20 \frac{1 - (\frac{3^{10}}{2^{10}})}{-\frac{1}{2}}$

This results in: