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 k² - 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 k² - 7k - 60 = 0. (b) Hence sh... 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 k² - 7k - 60 = 0.

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Answer

To show that the terms form a geometric series, we can set up the relationship between the terms. For a geometric series, the ratio between successive terms must be constant.

Let the first term be a = k + 4, and the second term be b = k. Then, the common ratio r can be expressed as:

$r = \frac{b}{a} = \frac{k}{k + 4}$

For the second and third terms:

$r = \frac{(2k - 15)}{k}$

Equating the two expressions for r gives us:

$\frac{k}{k + 4} = \frac{(2k - 15)}{k}$

Cross-multiplying leads to:

$k^2 = (k + 4)(2k - 15)$

Expanding the right-hand side:

$k^2 = 2k^2 - 15k + 8k - 60$ $k^2 = 2k^2 - 7k - 60$

Rearranging gives us:

$0 = 2k^2 - k^2 - 7k - 60$ $k^2 - 7k - 60 = 0.$

Step 2

Hence show that k = 12.

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Answer

To solve the quadratic equation derived:

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

We can factor it as:

$(k - 12)(k + 5) = 0$

This gives the solutions:

$k - 12 = 0 \Rightarrow k = 12,$ $k + 5 = 0 \Rightarrow k = -5.$

Since k is a positive constant, we have:

$k = 12.$

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:

Using the first term a:

$a = k + 4 = 12 + 4 = 16$

And the second term b:

$b = k = 12$

Using these expressions for the common ratio:

$r = \frac{b}{a} = \frac{12}{16} = \frac{3}{4}.$

Thus, the common ratio is: $r = 0.75$ .

Step 4

Find the sum to infinity of this series.

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Answer

The formula for the sum to infinity S of a geometric series is given by:

$S = \frac{a}{1 - r}$

For our series:

First term a = 16
Common ratio r = 0.75

Substituting these values into the formula gives:

$S = \frac{16}{1 - 0.75} = \frac{16}{0.25} = 64.$

Therefore, the sum to infinity of the series is: 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 k² - 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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