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 - 2017 - Paper 3

Given that k is not an integer, substituting into our derived equation:

$11k^2 - 130k + 99 = 0$

Applying the quadratic formula:

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

where a = 11, b = -130, and c = 99:

$k = \frac{130 \pm \sqrt{(-130)^2 - 4 \times 11 \times 99}}{2 \times 11}$

This simplifies to yield

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

which confirms our result.

Substituting k = \frac{9}{11} into the expression for the fourth term:

$a_4 = r^3 a_1$

First, find the common ratio r from the first term:

$r = \frac{5k - 7}{7k - 5} = \frac{5(\frac{9}{11}) - 7}{7(\frac{9}{11}) - 5} = \frac{ \frac{45}{11} - 7}{\frac{63}{11} - 5} = \frac{ \frac{45 - 77}{11}}{\frac{63 - 55}{11}} = \frac{-32}{8} = -4$

Thus, the fourth term is:

$a_4 = (-4)^3 (7(\frac{9}{11}) - 5) = -64(\frac{63 - 55}{11}) = -64(\frac{8}{11}) = \frac{-512}{11}$.

Using the formula for the sum of the first n terms of a geometric sequence:

$S_n = a_1 \frac{1 - r^n}{1 - r}$

Substituting the values we have:

$S_{10} = (7(\frac{9}{11}) - 5) \frac{1 - (-4)^{10}}{1 - (-4)}$

This calculates to:

$S_{10} = (\frac{63}{11} - 5)(\frac{1 - 1048576}{5}) = (\frac{8}{11})(\frac{-1048575}{5}) = \frac{-8396600}{55}$.

Thus, the sum of the first ten terms is approximately \frac{-8396600}{55}.