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

Oct 24, 2015

${\left(\frac{3}{4}\right)}^{14} \cdot 52 \approx 0.9265$

#### Explanation:

The first term is ${a}_{1} = 52$ and the common ratio is $\frac{3}{4}$:

$\frac{39}{52} = \frac{3 \cdot 13}{4 \cdot 13} = \frac{3}{4}$

$\frac{29.25}{39} = \frac{9.75 \cdot 3}{9.75 \cdot 4} = \frac{3}{4}$

So in general we can write:

${a}_{n} = {r}^{n - 1} {a}_{0} = {\left(\frac{3}{4}\right)}^{n} \cdot 52$

So we find:

${a}_{15} = {r}^{14} {a}_{0} = {\left(\frac{3}{4}\right)}^{14} \cdot 52 = \frac{{3}^{14} \cdot 52}{4} ^ 14$

$= \frac{4782969 \cdot 52}{268435456} = \frac{248714388}{268435456}$

$\approx 0.9265$