Question

What is the 15th term of the geometric sequence 52, 39, 29.25 …?

Answer 1

`(3/4)^14 * 52 ~~ 0.9265`

Explanation:

The first term is `a_1 = 52` and the common ratio is `3/4`:

`39/52 = (3*13)/(4*13) = 3/4`

`29.25/39 = (9.75*3)/(9.75*4) = 3/4`

So in general we can write:

`a_n = r^(n-1) a_0 = (3/4)^n*52`

So we find:

`a_15 = r^14 a_0 = (3/4)^14 * 52 = (3^14 * 52)/4^14`

`=(4782969 * 52) / 268435456 = 248714388 / 268435456`

`~~ 0.9265`