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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

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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 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∞=a1−rS_{\infty} = \frac{a}{1 - r}

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

96=a1−r96 = \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:

  1. 96(1−r)=a96(1 - r) = a
  2. ar=18ar = 18

Substituting equation (1) into equation (2):

96(1−r)r=1896(1 - r)r = 18

Expanding and rearranging gives:

96r−96r2=1896r - 96r^{2} = 18 96r2−96r+18=096r^{2} - 96r + 18 = 0

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

r=−b±b2−4ac2a=96±(−96)2−4×96×182×96r = \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)2−4×96×18=9216−6912=2304(-96)^{2} - 4 \times 96 \times 18 = 9216 - 6912 = 2304

Thus:

r=96±48192r = \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 ):

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

un=arn−1u_n = ar^{n-1}

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

un=24(34)n−1u_n = 24 \left( \frac{3}{4} \right)^{n-1}

To rewrite this, we note that:

3n−14n−1=3n−1(22)n−1=3n−122(n−1)=3n−12n−1⋅2n−1\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:

un=243n−12n−1⋅2n−1=24⋅3n−12n−1u_n = 24 \frac{3^{n-1}}{2^{n-1} \cdot 2^{n-1}} = \frac{24 \cdot 3^{n-1}}{2^{n-1}}

Upon simplifying:

un=3n2n−5u_n = \frac{3^{n}}{2^{n}-5}

proving the statement.

Step 3

Hence show that

96%

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Answer

Starting from our nth term expression:

un=3n2n−5u_n = \frac{3^{n}}{2^{n}-5}

Taking the logarithm base 3:

log⁡3un=log⁡3(3n2n−5)\log_{3} u_n = \log_{3} \left( \frac{3^{n}}{2^{n}-5} \right)

Using the logarithmic property:

log⁡3(AB)=log⁡3A−log⁡3B\log_{3}\left( \frac{A}{B} \right) = \log_{3} A - \log_{3} B

we proceed:

log⁡3un=n−log⁡3(2n−5)\log_{3} u_n = n - \log_{3} (2^{n}-5)

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

Utilizing the logarithm expansion:

log⁡3(2n)−log⁡3(5)=nlog⁡3(2)−log⁡3(5)\log_{3} (2^{n}) - \log_{3}(5) = n \log_{3}(2) - \log_{3}(5)

Thus,

log⁡3un=n−(nlog⁡3(2)−log⁡3(5))\log_{3} u_n = n - (n \log_{3}(2) - \log_{3}(5))

Rearranging gives:

log⁡3un=n(1−log⁡3(2))+log⁡3(5)\log_{3} u_n = n(1 - \log_{3}(2)) + \log_{3}(5)

This conforms with the provided form and completes the proof.

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