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.

A-Level Edexcel Maths Pure Quadratics: The first three terms of a geome

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 2 - 2009 - Paper 2

Question 2

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 2 - 2009 - Paper 2

Step 1

Show that k² - 7k - 60 = 0.

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Answer

To show that k² - 7k - 60 = 0, we start by using the fact that for a geometric series, the ratio of consecutive terms is constant. Therefore, we can set up the following equations:

Let the first term be: $a = k + 4$

The second term is: $ar = k$

The third term is: $ar^2 = 2k - 15$

From the first equation: $r = \frac{k}{k + 4}$

From the second equation: Substituting for r, we have: $2k - 15 = \frac{k(2k - 15)}{k + 4}$

Cross multiplying gives: $(2k - 15)(k + 4) = k(2k - 15)$ Expanding this, we get: $2k^2 + 8k - 15k - 60 = 2k^2 - 15k$ This simplifies to: $-7k - 60 = 0$ Thus, we arrive at the equation k² - 7k - 60 = 0.

Step 2

Hence show that k = 12.

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Answer

To solve the quadratic equation k² - 7k - 60 = 0, we can factor it: $(k - 12)(k + 5) = 0$ Therefore, the possible values of k are: $k - 12 = 0 \Rightarrow k = 12$ $k + 5 = 0 \Rightarrow k = -5$ Since k is a positive constant, we choose: $k = 12$

Step 3

Find the common ratio of this series.

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Answer

We now use k = 12 to find the common ratio (r): $r = \frac{k}{k + 4} = \frac{12}{12 + 4} = \frac{12}{16} = \frac{3}{4}$ Thus, the common ratio of the series is: $r = \frac{3}{4}$

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 with common ratio r where |r| < 1 is given by: $S = \frac{a}{1 - r}$ Substituting for a and r: $S = \frac{12 + 4}{1 - \frac{3}{4}} = \frac{16}{\frac{1}{4}} = 16 \times 4 = 64$ Therefore, 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 2 - 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 2 - 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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