The problem consists of two parts. The first part asks us to analyze the geometric sequence 64, 16, 4, ... and find (a) the common ratio, (b) the $n$th term, and (c) the sum to infinity. The second part asks us to find the geometric mean of 15 and 60.

1. Problem Description

The problem consists of two parts. The first part asks us to analyze the geometric sequence 64, 16, 4, ... and find (a) the common ratio, (b) the $n$th term, and (c) the sum to infinity. The second part asks us to find the geometric mean of 15 and

6

0.

2. Solution Steps

Part 1: Geometric Sequence 64, 16, 4, ...

(a) Common Ratio:

The common ratio $r$ is found by dividing any term by its preceding term.

$r = \frac{16}{64} = \frac{1}{4}$

$r = \frac{4}{16} = \frac{1}{4}$

So the common ratio $r = \frac{1}{4}$.

(b) $n$th Term:

The general formula for the $n$th term of a geometric sequence is:

$a_n = a_1 * r^{n-1}$

where $a_1$ is the first term and $r$ is the common ratio.

In our case, $a_1 = 64$ and $r = \frac{1}{4}$.

So, $a_n = 64 * (\frac{1}{4})^{n-1}$.

(c) Sum to Infinity:

The sum to infinity of a geometric sequence is given by the formula:

$S_\infty = \frac{a_1}{1 - r}$

provided that $|r| < 1$.

In our case, $a_1 = 64$ and $r = \frac{1}{4}$. Since $|\frac{1}{4}| < 1$, the sum to infinity exists.

$S_\infty = \frac{64}{1 - \frac{1}{4}} = \frac{64}{\frac{3}{4}} = 64 * \frac{4}{3} = \frac{256}{3}$.

Part 2: Geometric Mean of 15 and 60

The geometric mean of two numbers $a$ and $b$ is given by:

$GM = \sqrt{a * b}$

In this case, $a = 15$ and $b = 60$.

$GM = \sqrt{15 * 60} = \sqrt{900} = 30$.

3. Final Answer

1. (a) Common ratio: $\frac{1}{4}$

(b) $n$th term: $64 * (\frac{1}{4})^{n-1}$

(c) Sum to infinity: $\frac{256}{3}$

2. Geometric mean of 15 and 60: 30