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

Using the quadratic formula, we have: $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

a = 11, b = -130, c = 99;

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

Taking the square root gives: $\sqrt{12544} = 112$

Now, substituting back into the quadratic formula yields: $k = \frac{130 \pm 112}{22}$

This results in two solutions:

1. ( k = \frac{242}{22} = 11 ) (integer, rejected)
2. ( k = \frac{18}{22} = \frac{9}{11} ) (valid, as k is not an integer).

To find the fourth term (a_4), we need to use the formula for the nth term in a geometric sequence, given as:

$a_n = ar^{n-1}$

Where (a = 7k - 5) and (r = \frac{5k - 7}{7k - 5}$$. Substituting (k = \frac{9}{11}):

1. Calculate (a = 7\left(\frac{9}{11}\right) - 5 = \frac{63}{11} - \frac{55}{11} = \frac{8}{11})

2. Calculate (r = \frac{5\left(\frac{9}{11}\right) - 7}{7\left(\frac{9}{11}\right) - 5} = \frac{\frac{45}{11} - \frac{77}{11}}{\frac{63}{11} - \frac{55}{11}} = \frac{-32/11}{8/11} = -4)

Then:

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

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

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

Substituting the values:

• a = (\frac{8}{11})
• r = -4
• n = 10

Calculating yields:

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

Simplifying the parts:

• ((-4)^{10} = 1048576)
• Therefore:

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

• Hence:

$S_{10} = \frac{-8394200}{55} \approx -152520$