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We are given a geometric sequence \( a_1, a_2, a_3, \dots, a_n \), where we know the values:

\[
a_2 = \frac{4}{25}, \quad a_4 = \frac{16}{625}
\]
and the common ratio \( r > 0 \). We are asked to find the infinite sum:

\[
\sum_{n=1}^{\infty} a_n
\]

### Step 1: Determine the Common Ratio \( r \)

In a geometric sequence, the ratio between consecutive terms is constant. Therefore, we can use the known values of \( a_2 \) and \( a_4 \) to find \( r \). The general term of a geometric sequence is:

\[
a_n = a_1 \cdot r^{n-1}
\]

From the information given:

\[
a_2 = a_1 \cdot r = \frac{4}{25}
\]
\[
a_4 = a_1 \cdot r^3 = \frac{16}{625}
\]

Dividing \( a_4 \) by \( a_2 \) will allow us to solve for \( r^2 \):

\[
\frac{a_4}{a_2} = \frac{a_1 \cdot r^3}{a_1 \cdot r} = r^2 = \frac{\frac{16}{625}}{\frac{4}{25}} = \frac{16}{625} \times \frac{25}{4} = \frac{16 \times 25}{625 \times 4} = \frac{400}{2500} = \frac{1}{5}
\]

So, \( r^2 = \frac{1}{5} \), which gives:

\[
r = \frac{1}{\sqrt{5}}
\]

### Step 2: Solve for \( a_1 \)

Using the equation \( a_2 = a_1 \cdot r = \frac{4}{25} \) and substituting \( r = \frac{1}{\sqrt{5}} \), we can solve for \( a_1 \):

\[
a_1 \cdot \frac{1}{\sqrt{5}} = \frac{4}{25}
\]

\[
a_1 = \frac{4}{25} \cdot \sqrt{5} = \frac{4 \sqrt{5}}{25}
\]

### Step 3: Infinite Sum Formula

The sum of an infinite geometric series is given by:

\[
S = \frac{a_1}{1 - r} \quad \text{for} \ |r| < 1
\]

Substituting \( a_1 = \frac{4 \sqrt{5}}{25} \) and \( r = \frac{1}{\sqrt{5}} \):

\[
S = \frac{\frac{4 \sqrt{5}}{25}}{1 - \frac{1}{\sqrt{5}}} = \frac{\frac{4 \sqrt{5}}{25}}{\frac{\sqrt{5} - 1}{\sqrt{5}}} = \frac{4 \sqrt{5}}{25} \times \frac{\sqrt{5}}{\sqrt{5} - 1} = \frac{20}{25(\sqrt{5} - 1)}
\]

\[
S = \frac{4}{5(\sqrt{5} - 1)}
\]

This is the sum of the infinite series.

Would you like a more detailed explanation of any part of the process?

Here are five related questions you might want to explore:

1. How can we derive the formula for the sum of an infinite geometric series?
2. What is the condition for convergence of an infinite geometric series?
3. Can we solve for other terms of the sequence using the given information?
4. How would the process differ if the common ratio were negative?
5. How does the common ratio affect the behavior of the series as \( n \to \infty \)?

**Tip:** When solving geometric series problems, always verify that the common ratio satisfies \( |r| < 1 \) to ensure the sum converges.

Find the infinite sum of the given geometric sequence where a2 = 4/25 and a4 = 16/625 with r > 0.

Find the infinite sum of a geometric sequence where a2 = 4/25 and a4 = 16/625. This solution involves determining the common ratio and applying the infinite geometric series formula. Suitable for students in Grades 10-12.

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