Photo AI
A geometric series has first term 5 and common ratio \( \frac{4}{5} \) - Edexcel - A-Level Maths Pure - Question 7 - 2008 - Paper 2
Question 7
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 7 - 2008 - Paper 2
Step 1
a) the 20th term of the series, to 3 decimal places
Answer
To find the 20th term of a geometric series, we use the formula for the nth term:
[ T_n = a r^{n-1} ]
Where:
- ( a = 5 ) (the first term)
- ( r = \frac{4}{5} ) (the common ratio)
- ( n = 20 )
Plugging in the values:
[ T_{20} = 5 \left( \frac{4}{5} \right)^{19} = 5 \times \frac{4^{19}}{5^{19}} \approx 0.072 ]
Thus, the 20th term of the series is approximately ( 0.072 ) to three decimal places.
Step 2
b) the sum to infinity of the series
Answer
The sum to infinity for a geometric series is calculated using:
[ S_{\infty} = \frac{a}{1 - r} ]
The condition for this formula to be valid is that ( |r| < 1 ). Since ( r = \frac{4}{5} ), this condition is satisfied.
Substituting the values:
[ S_{\infty} = \frac{5}{1 - \frac{4}{5}} = \frac{5}{\frac{1}{5}} = 25 ]
Therefore, the sum to infinity of the series is ( 25 ).
Step 3
c) show that k > log 0.002 / log 0.8
Answer
Given the sum to ( k ) terms is greater than 24.95, we use the formula for the sum of the first ( k ) terms:
[ S_k = \frac{a(1 - r^k)}{1 - r} ]
Setting this greater than 24.95:
[ \frac{5(1 - r^k)}{1 - \frac{4}{5}} > 24.95 ]
This simplifies to:
[ 5(1 - r^k) > 24.95 ] [ 1 - r^k > 4.99 ] [ -r^k > 3.99 ] [ r^k < -3.99 ] (not valid as ( r^k ) cannot be negative)
Instead, rearranging gives:
[ r^k < 1 - \frac{24.95(1 - \frac{4}{5})}{5} ] [ r^k < 0.8 ]
Taking logs, we observe:
[ k \log r < \log 0.8 \quad \text{(if } r < 1\text{)} ] [ k > \frac{\log 0.002}{\log 0.8} ]
Thus, this shows that ( k > \frac{\log 0.002}{\log 0.8} ).
Step 4
d) find the smallest possible value of k
Answer
From part (c), we derived:
[ k > \frac{\log 0.002}{\log 0.8} ]
Calculating this value:
- ( \log 0.002 \approx -2.6990 )
- ( \log 0.8 \approx -0.09691 )
Thus:
[ k > \frac{-2.6990}{-0.09691} \approx 27.8 ]
The smallest integer value of ( k ) that satisfies this condition is ( k = 28 ).