We are given:

• The first term $a = 125$
• The last term $l = \frac{1}{5}$
• The common ratio $r = \frac{1}{5}$

Step 1: Determine the Number of Terms

For a geometric sequence, the last term $l$ is given by:

$l = a \cdot r^{n-1}$

Substituting the given values:

$\frac{1}{5} = 125 \cdot \left(\frac{1}{5}\right)^{n-1}$

Step 2: Solve for $n$

First, divide both sides by 125:

$\frac{1}{5} \div 125 = \left(\frac{1}{5}\right)^{n-1}$

$\frac{1}{5} \times \frac{1}{125} = \left(\frac{1}{5}\right)^{n-1}$

$\frac{1}{625} = \left(\frac{1}{5}\right)^{n-1}$

Since $\frac{1}{625} = \left(\frac{1}{5}\right)^4$, we have:

$n - 1 = 4 \implies n = 5$

So, there are 5 terms in the sequence.

Step 3: Calculate the Sum

The formula for the sum $S_n$ of a geometric sequence is:

$S_n = a \cdot \frac{1 - r^n}{1 - r} \quad \text{if} \ r \neq 1$

Substituting the known values:

$S_5 = 125 \cdot \frac{1 - \left(\frac{1}{5}\right)^5}{1 - \frac{1}{5}}$

Step 4: Simplify the Expression

First, calculate $\left(\frac{1}{5}\right)^5$:

$\left(\frac{1}{5}\right)^5 = \frac{1}{3125}$

So the sum becomes:

$S_5 = 125 \cdot \frac{1 - \frac{1}{3125}}{\frac{4}{5}}$

$S_5 = 125 \cdot \frac{\frac{3125 - 1}{3125}}{\frac{4}{5}}$

$S_5 = 125 \cdot \frac{\frac{3124}{3125}}{\frac{4}{5}}$

$S_5 = 125 \cdot \frac{3124}{3125} \cdot \frac{5}{4}$

$S_5 = 125 \cdot \frac{15620}{12500}$

$S_5 = 125 \cdot 1.2496$

$S_5 \approx 156.2$

Final Answer

The sum of the sequence is approximately 156.2.

Would you like more details or have any other questions?

Here are five related questions:

1. How do you determine the last term if the number of terms is unknown?
2. How does the sum change if the common ratio is greater than 1?
3. Can you calculate the sum of an infinite geometric series when the common ratio is a fraction?
4. What is the formula for the nth term in a geometric series?
5. How does the common ratio affect the growth or decay of a geometric sequence?

Tip: Always double-check the value of the common ratio when dealing with geometric series problems to avoid calculation errors.