The first three terms of a geometric series are 4p, (3p + 15) and (5p + 20) respectively, where p is a positive constant. (a) Show that 11p² - 10p - 225 = 0 (b) Hence show that p = 5 (c) Find the common ratio of this series. (d) Find the sum of the first ten terms of the series, giving your answer to the nearest integer.

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

The first three terms of a geometric series are 4p, (3p + 15) and (5p + 20) respectively, where p is a positive constant - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 5

Question 7

The first three terms of a geometric series are 4p, (3p + 15) and (5p + 20) respectively, where p is a positive constant. (a) Show that 11p² - 10p - 225 = 0 (b) He... show full transcript

Worked Solution & Example Answer:The first three terms of a geometric series are 4p, (3p + 15) and (5p + 20) respectively, where p is a positive constant - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 5

Step 1

Show that 11p² - 10p - 225 = 0

96%

114 rated

Answer

Next, we can solve the quadratic equation $11p^2 - 10p - 225 = 0$ using the quadratic formula:

$p = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

Where a = 11, b = -10, c = -225. Calculating the discriminant:\n

Discriminant: $(-10)^2 - 4 \times 11 \times (-225) = 100 + 9900 = 10000$

Now substituting into the formula:

$p = \frac{-(-10) \pm 100}{2 \times 11} = \frac{10 \pm 100}{22}$

This gives us:

$p = \frac{110}{22} = 5$ (valid since p is positive)
$p = \frac{-90}{22}$ (not valid as p must be positive)

Therefore, we conclude that ( p = 5 ).

Step 2

Find the common ratio of this series.

99%

104 rated

Answer

To find the common ratio (r) of the geometric series, we can use the first two terms:

$r = \frac{a_2}{a_1} = \frac{3p + 15}{4p}$

Substituting p = 5:

$r = \frac{3(5) + 15}{4(5)} = \frac{30}{20} = \frac{3}{2}$

Thus, the common ratio is ( r = \frac{3}{2} ).

Step 3

Find the sum of the first ten terms of the series, giving your answer to the nearest integer.

96%

101 rated

Answer

The formula for the sum S of the first n terms of a geometric series is given by:

$S_n = \frac{a(1 - r^n)}{1 - r}$

Here, ( a = 4p ) and ( r = \frac{3}{2} ), substituting p = 5:

$a = 4(5) = 20 \quad\text{and}\quad n = 10$

Substituting these values into the sum formula:

$S_{10} = \frac{20(1 - (\frac{3}{2})^{10})}{1 - \frac{3}{2}} = \frac{20(1 - \frac{3^{10}}{2^{10}})}{-\frac{1}{2}} = -40(1 - \frac{59049}{1024})$

Calculating the above gives approximately ( S_{10} \approx 2267 ) after rounding to the nearest integer.

The first three terms of a geometric series are 4p, (3p + 15) and (5p + 20) respectively, where p is a positive constant - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 5

Worked Solution & Example Answer:The first three terms of a geometric series are 4p, (3p + 15) and (5p + 20) respectively, where p is a positive constant - Edexcel - A-Level Maths Pure - Question 7 - 2013 - Paper 5

Show that 11p² - 10p - 225 = 0

Find the common ratio of this series.

Find the sum of the first ten terms of the series, giving your answer to the nearest integer.

Join the A-Level students using SimpleStudy...