The first three terms of a geometric sequence are $7k - 5$, $5k - 7$, $2k + 10$ where $k$ is a constant - Edexcel - A-Level Maths Pure - Question 2 - 2016 - Paper 2

To find the values of $k$, we use the quadratic formula on the derived equation:

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

Where $a = 11$, $b = -130$, and $c = 99$.

Calculating the discriminant:

$b^2 - 4ac = (-130)^2 - 4(11)(99) = 16900 - 4356 = 12544$

Thus, the roots are:

$k = \frac{130 \pm \sqrt{12544}}{22}$

This simplifies to:

$k = \frac{130 \pm 112}{22}$

Calculating both potential solutions gives:

$k_1 = \frac{242}{22} = 11$ $k_2 = \frac{18}{22} = \frac{9}{11}$

Since $k$ is not an integer, we conclude that:

$k = \frac{9}{11}$.

To find the fourth term of the sequence, we can use the formula for the nth term of a geometric sequence:

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

We first need to find the common ratio $r$. Using the values of $k = \frac{9}{11}$:

• First term: $a = 7 \left( \frac{9}{11} \right) - 5 = \frac{63}{11} - \frac{55}{11} = \frac{8}{11}$
• Second term: $ar = 5 \left( \frac{9}{11} \right) - 7 = \frac{45}{11} - \frac{77}{11} = -\frac{32}{11}$

The common ratio $r$ is:

$r = \frac{-\frac{32}{11}}{\frac{8}{11}} = -4$

Now calculating the fourth term:

$a_4 = ar^{3} = \frac{8}{11}(-4)^{3} = \frac{8}{11}(-64) = -\frac{512}{11}$.

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

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

Using $a = -\frac{512}{11}$, $r = -4$, and $n = 10$:

$S_{10} = -\frac{512}{11} \frac{1 - (-4)^{10}}{1 - (-4)}$

Calculating $(-4)^{10} = 1048576$ gives:

$S_{10} = -\frac{512}{11} \frac{1 - 1048576}{1 + 4} = -\frac{512}{11} \frac{-1048575}{5}$

Now simplifying:

$S_{10} = \frac{512 \times 1048575}{55} = -152520.90909...$

Thus,

$S_{10} = -152520.90909...$

This is the sum of the first ten terms.