## Precalculus (6th Edition) Blitzer

a. See table and explanations. b. $a_n=15.98(1.02)^{n-1}$ c. $28.95$ (million)
a. See table. The first column is the year, the second column is the population (in millions), the third column is the ratio of the population divided by that of the preceding year, and the last column is the ratio rounded to two decimal places. We can see that $r=1.02$, indicating that the population increase approximately geometrically. b. Let $n$ be the year after 1999. We can write the general term of the geometric sequence as $a_n=a_1r^{n-1}=15.98(1.02)^{n-1}$ c. For year 2030, we have $n=2030-1999=31$. Thus $a_{31}=15.98(1.02)^{31-1}\approx28.95$ (million)