Worked Solution:
The sum, S of a geometric series of n terms | starting at a with common ratio | r is given by S = | | a (1 – rn) |  | | 1 – r |
|
The questions asks for
n:
| rn = 1 – | | S (1 – r) | | | a |
|
Taking logs of both sides:
| n log r = log(1 – | | S (1 – r) | | | a |
| ) |
Putting the values in the question into this formula we have:
| n log | | 11 | | | 5 |
| = log | ( | 1 – | | ) |
n log 2.2 = log 270001
Using logs to base 10, the number of terms to make a sum of
600000 is
| n = | | log10 270001 | | | log10 2.2 |
| = | | 5.431365 | | | 0.342423 |
| = 15.862 |
Since n must be a whole number, the sum is first exceeded when
n = 16
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