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A geometric series has first term 5 and common ratio \( \frac{4}{5} \) - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 2
Question 8
A geometric series has first term 5 and common ratio \( \frac{4}{5} \). Calculate (a) the 20th term of the series, to 3 decimal places, (b) the sum to infinity of... show full transcript
Worked Solution & Example Answer:A geometric series has first term 5 and common ratio \( \frac{4}{5} \) - Edexcel - A-Level Maths Pure - Question 8 - 2008 - Paper 2
Step 1
the 20th term of the series, to 3 decimal places
Answer
To find the 20th term ( T_{20} ), we use the formula for the ( n )-th term of a geometric series:
[ T_n = a r^{n-1} ]
where ( a = 5 ) (the first term) and ( r = \frac{4}{5} ) (the common ratio). For ( n = 20 ):
[ T_{20} = 5 \left( \frac{4}{5} \right)^{19} ]
Calculating this gives:
[ T_{20} = 5 \times \frac{4^{19}}{5^{19}} = \frac{5 \times 4^{19}}{5^{19}} = 5 \times \left( \frac{4}{5} \right)^{19} \approx 0.072 ]
Thus, the 20th term is approximately 0.072.
Step 2
the sum to infinity of the series
Answer
The sum to infinity ( S ) of a geometric series is given by:
[ S = \frac{a}{1 - r} ]
For our series, substituting ( a = 5 ) and ( r = \frac{4}{5} ):
[ S \approx \frac{5}{1 - \frac{4}{5}} = \frac{5}{\frac{1}{5}} = 25 ]
Therefore, the sum to infinity is 25.
Step 3
show that k > log 0.002 / log 0.8
Answer
The formula for the sum of the first ( k ) terms of a geometric series is:
[ S_k = \frac{a(1 - r^k)}{1 - r} ]
We want to find ( k ) such that:
[ \frac{5(1 - (\frac{4}{5})^k)}{1 - \frac{4}{5}} > 24.95 ]
This simplifies to:
[ 5(1 - (\frac{4}{5})^k) > 24.95 \times \frac{1}{5} ]
Then:
[ 1 - (\frac{4}{5})^k > 4.99 ]
This gives:
[ (\frac{4}{5})^k < 1 - 4.99 \Rightarrow (\frac{4}{5})^k < 0.001\ ]
By taking logarithms on both sides:
[ k \log(\frac{4}{5}) < \log(0.001) ]
Since ( \log(\frac{4}{5}) < 0 ), we reverse the inequality:
[ k > \frac{\log(0.002)}{\log(0.8)} ]
Step 4