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The sum to infinity of a geometric series is 96 The first term of the series is less than 30 The second term of the series is 18 8 (a) Find the first term and common ratio of the series - AQA - A-Level Maths Pure - Question 8 - 2020 - Paper 3

Question 8

The sum to infinity of a geometric series is 96 The first term of the series is less than 30 The second term of the series is 18 8 (a) Find the first term and commo... show full transcript

Worked Solution & Example Answer:The sum to infinity of a geometric series is 96 The first term of the series is less than 30 The second term of the series is 18 8 (a) Find the first term and common ratio of the series - AQA - A-Level Maths Pure - Question 8 - 2020 - Paper 3

Step 1

Find the first term and common ratio of the series.

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Answer

To find the first term and the common ratio of the geometric series, we start by using the formula for the sum to infinity:

$S_{\infty} = \frac{a}{1 - r}$

where ( S_{\infty} = 96 ). Thus, we have the equation:

$96 = \frac{a}{1 - r}$

From the information given, we also know:

The first term ( a < 30 )
The second term is given by ( ar = 18 )

Now we have a system of equations:

$96(1 - r) = a$
$ar = 18$

Substituting equation (1) into equation (2):

$96(1 - r)r = 18$

Expanding and rearranging gives:

$96r - 96r^{2} = 18$ $96r^{2} - 96r + 18 = 0$

Now we can use the quadratic formula to solve for r:

$r = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a} = \frac{96 \pm \sqrt{(-96)^{2} - 4 \times 96 \times 18}}{2 \times 96}$

Calculating the discriminant:

$(-96)^{2} - 4 \times 96 \times 18 = 9216 - 6912 = 2304$

Thus:

$r = \frac{96 \pm 48}{192}$

This yields two possible values:

( r = \frac{24}{48} = \frac{1}{2} )
( r = \frac{144}{192} = \frac{3}{4} )

Now substituting back to find ( a ) for each ( r ):

If ( r = \frac{1}{2} ): ( a = 96(1 - \frac{1}{2}) = 48 ) (not valid since ( a < 30 ))
If ( r = \frac{3}{4} ): ( a = 96(1 - \frac{3}{4}) = 24 )

Thus, the first term ( a = 24 ) and the common ratio ( r = \frac{3}{4} ).

Step 2

Show that the nth term of the series, u_n, can be written as

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Answer

The nth term of a geometric series is given by:

$u_n = ar^{n-1}$

Substituting our values for ( a ) and ( r ):

$u_n = 24 \left( \frac{3}{4} \right)^{n-1}$

To rewrite this, we note that:

$\frac{3^{n-1}}{4^{n-1}} = \frac{3^{n-1}}{(2^{2})^{n-1}} = \frac{3^{n-1}}{2^{2(n-1)}} = \frac{3^{n-1}}{2^{n-1} \cdot 2^{n-1}}$

Thus:

$u_n = 24 \frac{3^{n-1}}{2^{n-1} \cdot 2^{n-1}} = \frac{24 \cdot 3^{n-1}}{2^{n-1}}$

Upon simplifying:

$u_n = \frac{3^{n}}{2^{n}-5}$

proving the statement.

Step 3

Hence show that

96%

101 rated

Answer

Starting from our nth term expression:

$u_n = \frac{3^{n}}{2^{n}-5}$

Taking the logarithm base 3:

$\log_{3} u_n = \log_{3} \left( \frac{3^{n}}{2^{n}-5} \right)$

Using the logarithmic property:

$\log_{3}\left( \frac{A}{B} \right) = \log_{3} A - \log_{3} B$

we proceed:

$\log_{3} u_n = n - \log_{3} (2^{n}-5)$

Now focusing on ( \log_{3} (2^{n}-5) ):

Utilizing the logarithm expansion:

$\log_{3} (2^{n}) - \log_{3}(5) = n \log_{3}(2) - \log_{3}(5)$

Thus,

$\log_{3} u_n = n - (n \log_{3}(2) - \log_{3}(5))$

Rearranging gives:

$\log_{3} u_n = n(1 - \log_{3}(2)) + \log_{3}(5)$

This conforms with the provided form and completes the proof.

The sum to infinity of a geometric series is 96 The first term of the series is less than 30 The second term of the series is 18 8 (a) Find the first term and common ratio of the series - AQA - A-Level Maths Pure - Question 8 - 2020 - Paper 3

Worked Solution & Example Answer:The sum to infinity of a geometric series is 96 The first term of the series is less than 30 The second term of the series is 18 8 (a) Find the first term and common ratio of the series - AQA - A-Level Maths Pure - Question 8 - 2020 - Paper 3

Find the first term and common ratio of the series.

Show that the nth term of the series, u_n, can be written as

Hence show that

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