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 - 2017 - Paper 3

Given ( k ) is not an integer, we proceed to solve the quadratic equation using the quadratic formula:

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

Substituting ( a = 11, b = -130, c = 99 ):

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

Calculate the discriminant:

$16900 - 4356 = 12544$

Now find ( k ):

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

This gives two possible solutions:

1. ( k = \frac{242}{22} = 11 ) (not valid since we are given that k is not an integer)
2. ( k = \frac{18}{22} = \frac{9}{11} ) (valid solution)

For ( k = \frac{9}{11} ), substituting into the expression for the first three terms:

1. First term: ( 7k - 5 = 7(\frac{9}{11}) - 5 = \frac{63}{11} - \frac{55}{11} = \frac{8}{11} )
2. Second term: ( 5k - 7 = 5(\frac{9}{11}) - 7 = \frac{45}{11} - \frac{77}{11} = -\frac{32}{11} )
3. Third term: ( 2k + 10 = 2(\frac{9}{11}) + 10 = \frac{18}{11} + \frac{110}{11} = \frac{128}{11} )

Now, find the fourth term using the common ratio ( r ):

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

Thus, the fourth term is:

$\text{Fourth term} = \text{Third term} \times r = \frac{128}{11} \times -4 = -\frac{512}{11}$

The formula for the sum of the first ( n ) terms of a geometric series is:

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

where ( a ) is the first term and ( r ) is the common ratio. From prior calculations:

1. ( a = \frac{8}{11} )
2. ( r = -4 )

For ( n = 10 ):

$S_{10} = \frac{8}{11} \frac{1 - (-4)^{10}}{1 - (-4)}$

Calculating further:

((-4)^{10} = 1048576)

So:

$S_{10} = \frac{8}{11} \frac{1 - 1048576}{1 + 4} = \frac{8}{11} \frac{-1048575}{5} = -\frac{8390600}{55}$