Answer to Question #349858 in Algebra for mengal

Question

A species of fish grows 1.2 cm in first year. Each year it grows by 2% more than it did the year before. Using geometric sequence find, how much does it grow in 10^th year.

Answer 1

Each year the growth is 2%=0.02 more than the previous year, so we have the geometric secuence with the first term 1.2 and the common ratio $r=1+0.02=1.02.$

We know that for geometric secuence $a_n=a_1\times r^{n-1},$

so in 10th year this species of fish grows

$a_{10}=1.2\times 1.02^9=1.4341110823467733 \approx 1.43$ cm.

Answer: 1.43 cm