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 10 - 2016 - Paper 2

To find the value of $k$, we solve the quadratic equation: [ 11k^2 - 130k + 99 = 0 ]

Using the quadratic formula: [ k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]

Where $a = 11$, $b = -130$, and $c = 99$. Thus, [ k = \frac{130 \pm \sqrt{(-130)^2 - 4 \cdot 11 \cdot 99}}{2 \cdot 11} ] [ = \frac{130 \pm \sqrt{16900 - 4356}}{22} ] [ = \frac{130 \pm \sqrt{15544}}{22} ] [ = \frac{130 , \pm , 124}{22} ]

This gives two potential solutions for $k$: one positive and one negative. As per the requirement, after solving we find that when $k = \frac{9}{11}$, it is not an integer.

With $k = \frac{9}{11}$, substitute into the expression for the fourth term: [ ar^3 = 7k - 5 \cdot r^3 ]

First, compute the common ratio $r$: [ r = \frac{5k - 7}{7k - 5} = \frac{5(\frac{9}{11}) - 7}{7(\frac{9}{11}) - 5} = \frac{\frac{45}{11} - \frac{77}{11}}{\frac{63}{11} - \frac{55}{11}} = \frac{-32/11}{8/11} = -4 ]

Then, we substitute: [ ar^3 = (7(\frac{9}{11}) - 5)(-4)^3 = (\frac{63}{11} - 5)(-64) ] [ = (\frac{63}{11} - \frac{55}{11})(-64) = (\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 $a = 7k - 5$ and $r = -4$, we compute: [ S_{10} = (7(\frac{9}{11}) - 5) \frac{1 - (-4)^{10}}{1 - (-4)} ] [ = (\frac{63}{11} - \frac{55}{11}) \frac{1 - 1048576}{5} ] [ = (\frac{8}{11}) \frac{-1048575}{5} = -\frac{8388600}{55} ]

Therefore, the sum of the first ten terms is: [ S_{10} = -152520.9090909090 ]