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

Now, we know that: $11k^2 - 130k + 99 = 0$

Using the quadratic formula,( k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ) where ( a = 11, b = -130, c = 99 ):

Calculate the discriminant:

• Discriminant = $(-130)^2 - 4(11)(99) = 16900 - 4356 = 15544$

Finding the roots: $k = \frac{130 \pm \sqrt{15544}}{22}$ Extracting the square root gives: $k = \frac{130 \pm 124}{22}$ Thus:

• First root: ( k = \frac{254}{22} = 11.5454 ) (not an integer)
• Second root: ( k = \frac{6}{22} = \frac{9}{11} ) (is a valid solution)

To find the fourth term, we first determine the common ratio (r) of the geometric sequence:

Using the first two terms: $r = \frac{5k - 7}{7k - 5}$
Substituting ( k = \frac{9}{11} ): $r = \frac{5(9/11) - 7}{7(9/11) - 5} = \frac{\frac{45}{11} - \frac{77}{11}}{\frac{63}{11} - \frac{55}{11}} = \frac{-\frac{32}{11}}{\frac{8}{11}} = -4$

Now find the fourth term ( a r^3 ): First term ( a = 7(\frac{9}{11}) - 5 = \frac{63}{11} - \frac{55}{11} = \frac{8}{11} )

Then the fourth term is: $\frac{8}{11}(-4)^3 = \frac{8}{11}(-64) = -\frac{512}{11}$

The sum of the first n terms of a geometric sequence is given by: $S_n = a \frac{1 - r^n}{1 - r}$

Using:

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

Plugging in the values: $S_{10} = \frac{8}{11} \frac{1 - (-4)^{10}}{1 - (-4)} = \frac{8}{11} \frac{1 - 1048576}{5}$ Calculating:

• ( (-4)^{10} = 1048576 ) gives: $S_{10} = \frac{8}{11} \frac{-1048575}{5} = -\frac{8388600}{55}$