## Karyatou Group Title The first three terms of a sequence are given. 15,10,20/3,... Find the 6th term. one year ago one year ago

1. across Group Title

Let $$a_0=15$$. Then$a_{n+1}=\frac{2a_n}{3}\qquad n=0,1,\dots$

2. across Group Title

Here, you're looking for $$a_5$$.

3. across Group Title

That is,$a_5=(2/3)(2/3)(2/3)(2/3)(2/3)a_0=\frac{32}{243}a_0=\frac{480}{243}\approx1.975308641975309.$