Question

The population of a town is 50,000 people in the year 2000. How many people with live in the town in 2009 if the population increases at a rate of 3% each year? Round your answer to the nearest whole number.

Answer 1

3^9 I think?
@lgbasallote

Answer 2

use geometric sequence
\[a_n = a_1 r^{n-1}\]
in this problem you're looking for \(a_{9}\) so there are 9 terms. then your common ratio would be 0.03 because it's 3% changed to decimal. then your first term is 50,000

\[a_n = 50,000 \times 0.03^{9-1}\]
you were close heh