# Find the smallest value of n for which the nth term of the series is less than 0.001

Question:

The common ratio and the first term of a geometric series are 0.55 and 18 respectively.

Find the smallest value of n for which the nth term of the series is less than 0.001

$${\text{My solution: }}$$

$$Tn<0.001$$ $$ar^{n-1}<0.001$$

$$(18)(0.55)^{n-1}<0.001$$

$$(0.55)^{n-1}<\frac{1}{18000}$$ $$(n-l){\log 0.55}<{\log \frac{1}{18000}}$$ $$n-1<16.389$$ $$n<16.389+1$$ $$n<17.389$$ $$n=17$$

$${\text{However, the answer given for this question is}}$$ $$n=18$$ $${\text{Would anyone tell me either it is my answer or the answer given that is wrong ,please?}}$$ $${\text{Thank you.}}$$

## 1 Answer

$\log(0.55) < 0$ so from $$(n-1) \, \log(0.55) \lt \log \frac{1}{18000}$$ dividing (or multiplying) both sides by a negative number means your next line should be $$n-1 \gt 16.389\ldots$$ changing the direction of the inequality.

Checking the answers, $18 \times 0.55^{17-1} = 0.001262\ldots$ while $18 \times 0.55^{18-1} = 0.000694\ldots$