C2 2017 9 Edexcel

9

(a)

As the sequence is geometric, the ratios of consecutive terms should be equal, that is:

\displaystyle \frac {5k - 7} {7k - 5} = \frac {2k + 10} {5k - 7}

So:

(7k - 5)(2k + 10) = (5k - 7)^2 = 25k^2 - 70k + 49

Expanding the LHS we have:

(7k - 5)(2k + 10) = 14k^2 - 10k + 70k - 50 = 14k^2 + 60k - 50

So:

25k^2 - 70k + 49 = 14k^2 + 60k - 50

So:

11k^2 - 130k + 99 = 0

(b)

We can factorise this to:

(11k - 9)(k - 11) = 0

So either k = \dfrac 9 {11} or k = 11. Since k is not an integer, we have k = \dfrac 9 {11}.

(ci)

Then the first term of the sequence is:

\displaystyle 7 \left(\frac 9 {11}\right) - 5 = \frac 8 {11}

and the common ratio is:

\displaystyle \frac {5 \left(\frac 9 {11}\right) - 7} {7 \left(\frac 9 {11}\right) - 5} = -4

So the fourth term in the sequence is:

\displaystyle \frac 8 {11} \times \left(-4\right)^3 = -\dfrac {512} {11}

(ii)

The sum of the first ten terms of the sequence is:

\displaystyle \frac 8 {11} \times \frac {\left(-4\right)^{10} - 1} {-4 - 1} = -152520

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