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The fourth term of a geometric series is 10 and the seventh term of the series is 80 - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 2
Question 4
The fourth term of a geometric series is 10 and the seventh term of the series is 80. For this series, find a) the common ratio, b) the first term, c) the sum of... show full transcript
Worked Solution & Example Answer:The fourth term of a geometric series is 10 and the seventh term of the series is 80 - Edexcel - A-Level Maths Pure - Question 4 - 2008 - Paper 2
Step 1
a) the common ratio
Answer
To find the common ratio, let the first term of the geometric series be denoted as ( a ) and the common ratio as ( r ). The nth term of a geometric series can be described by the formula:
[ T_n = a \cdot r^{n-1} ]
For the fourth term: [ T_4 = a \cdot r^{3} = 10\ ag{1} ]
For the seventh term: [ T_7 = a \cdot r^{6} = 80\ ag{2} ]
Now, divide equation (2) by equation (1):
[
\frac{T_7}{T_4} = \frac{a \cdot r^{6}}{a \cdot r^{3}} = \frac{80}{10}
]
This simplifies to:
[
r^{3} = 8
]
Therefore, the common ratio is:
[
r = 8^{\frac{1}{3}} = 2
]
Step 2
Step 3
c) the sum of the first 20 terms, giving your answer to the nearest whole number
Answer
The sum ( S_n ) of the first ( n ) terms of a geometric series is given by: [ S_n = a \frac{1 - r^{n}}{1 - r} ]
In this instance, for 20 terms ( ( n = 20 )), we have: [ S_{20} = 1.25 \frac{1 - 2^{20}}{1 - 2} ]
Calculating this: [ S_{20} = 1.25 \frac{1 - 1048576}{-1} = 1.25 \cdot (1048575) ]
This results in:
[
S_{20} \approx 1310718.75
]
Rounding to the nearest whole number gives:
[
S_{20} \approx 1310719
]