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
$${\text{My solution: }}$$
$$Tn<0.001$$
$$ar^{n-1}<0.001$$
$$(18)(0.55)^{n-1}<0.001$$
$$(0.55)^{n-1}<\frac{1}{18000}$$
$$(n-l){\log 0.55}<{\log \frac{1}{18000}}$$
$$n-1<16.389$$
$$n<16.389+1$$
$$n<17.389$$
$$n=17$$
$${\text{However, the answer given for this question is}}$$ 
$$n=18$$
$${\text{Would anyone tell me either it is my answer  or the answer given that is wrong ,please?}}$$ 
$${\text{Thank you.}}$$ 

$\log(0.55) < 0$ 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 $18 \times 0.55^{18-1} = 0.000694\ldots$
