Question

The first three terms of a sequence are given.
15,10,20/3,...
Find the 6th term.

Answer 1

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

Answer 2

Here, you're looking for \(a_5\).

Answer 3

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