# A geometric sequence has first term $256$ and ratio $0.75$. Find the smallest $n$ for which the sum of the first $n$ terms exceeds $1000$.

A geometric sequence whose first term = 256 and whose common ratio is 0.75. Find the smallest number of n for which the sum of the first n terms of the sequence exceeds 1000.

My turn: $$S_n = \frac{256(1-(0.75)^n)}{1-0.75} > 1000$$ $$n \log{0.75} < \log{\frac{3}{128}}$$ $$n < 13.04$$ then $$n = 13$$

What is wrong with my solution because 13 does not satisfy the requires but 14 does ?

• $\log(0.75)$ is negative. You forgot to swap the direction of the inequality sign. – JMoravitz Jul 23 at 20:39

Hint: For $$a,b>0$$, if $$c>0$$ then $$ac < b c \implies a < b.$$ What happens if $$c <0$$?