The first three terms of a geometric series are (k + 4), k and (2k - 15) respectively, where k is a positive constant. (a) Show that $k^2 - 7k - 60 = 0$. (b) Hence show that $k = 12$. (c) Find the common ratio of this series. (d) Find the sum to infinity of this series.

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The first three terms of a geometric series are (k + 4), k and (2k - 15) respectively, where k is a positive constant - Edexcel - A-Level Maths Pure - Question 10 - 2009 - Paper 2

Question 10

The first three terms of a geometric series are (k + 4), k and (2k - 15) respectively, where k is a positive constant. (a) Show that $k^2 - 7k - 60 = 0$. (b) Henc... show full transcript

Worked Solution & Example Answer:The first three terms of a geometric series are (k + 4), k and (2k - 15) respectively, where k is a positive constant - Edexcel - A-Level Maths Pure - Question 10 - 2009 - Paper 2

Step 1

Show that $k^2 - 7k - 60 = 0$

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Answer

To show that the terms form a geometric series, we can use the property that the ratio of successive terms is constant:

The first term is $a = k + 4$ .
The second term is $ar = k$ .
The third term is $ar^2 = 2k - 15$ .

From this, we can express the common ratio $r$ as: $r = \frac{k}{k+4} = \frac{2k - 15}{k}$

Cross-multiplying gives: $(k)(k) = (k + 4)(2k - 15)$

This simplifies to: $k^2 = (2k^2 - 15k + 8k - 60)$ $k^2 = 2k^2 - 7k - 60$

Rearranging this leads us to: $0 = k^2 - 7k - 60$

Thus, we have shown that $k^2 - 7k - 60 = 0$ .

Step 2

Hence show that $k = 12$

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To find the values of $k$ , we can factor or use the quadratic formula on the equation $k^2 - 7k - 60 = 0$ .

Factoring the quadratic: $(k - 12)(k + 5) = 0$

Setting each factor to zero gives: $k - 12 = 0 \Rightarrow k = 12$
$k + 5 = 0 \Rightarrow k = -5$

Since $k$ is a positive constant, we conclude that: $k = 12$ .

Step 3

Find the common ratio of this series.

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Answer

Now that we have $k = 12$ , we can find the common ratio $r$ .

Substituting $k$ into the formula for $r$ : $r = \frac{k}{k + 4} = \frac{12}{12 + 4} = \frac{12}{16} = \frac{3}{4}$

Thus, the common ratio is: $r = \frac{3}{4}$ .

Step 4

Find the sum to infinity of this series.

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Answer

The sum to infinity of a geometric series is given by: $S = \frac{a}{1 - r}$ where:

$a$ is the first term
$r$ is the common ratio.

Here, the first term is: $a = k + 4 = 12 + 4 = 16$ And the common ratio we found is: $r = \frac{3}{4}$

Substituting these values into the formula: $S = \frac{16}{1 - \frac{3}{4}} = \frac{16}{\frac{1}{4}} = 16 \times 4 = 64$

Thus, the sum to infinity of this series is: $S = 64$ .

The first three terms of a geometric series are (k + 4), k and (2k - 15) respectively, where k is a positive constant - Edexcel - A-Level Maths Pure - Question 10 - 2009 - Paper 2

Worked Solution & Example Answer:The first three terms of a geometric series are (k + 4), k and (2k - 15) respectively, where k is a positive constant - Edexcel - A-Level Maths Pure - Question 10 - 2009 - Paper 2

Show that $k^2 - 7k - 60 = 0$

Hence show that $k = 12$

Find the common ratio of this series.

Find the sum to infinity of this series.

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