In a geometric sequence, the sum of the first $n$ terms is given by the formula:

$S_n = a \frac{1 - r^n}{1 - r}$

Where:

• $S_n$ is the sum of the first $n$ terms,
• $a$ is the first term,
• $r$ is the common ratio,
• $n$ is the number of terms.

We are given:

• $S_5 = 121$ (the sum of the first 5 terms),
• $a = 1$ (the first term),
• $n = 5$ (the number of terms).

Substituting the values into the sum formula:

$121 = 1 \cdot \frac{1 - r^5}{1 - r}$

This simplifies to:

$121 = \frac{1 - r^5}{1 - r}$

Now, we need to solve for $r$. Let's go step by step.

First, multiply both sides by $1 - r$:

$121(1 - r) = 1 - r^5$

Expanding the left-hand side:

$121 - 121r = 1 - r^5$

Now, rearrange the terms:

$121 - 1 = 121r - r^5$

$120 = r^5 - 121r$

We now need to solve this equation for $r$. I'll calculate the value of $r$.The real solutions to the equation $r^5 - 121r - 120 = 0$ are $r = -3$ and $r = -1$. However, since geometric sequences typically deal with positive ratios (unless otherwise specified), the common ratio in this context is likely $r = -3$.

So, the common ratio is $\boxed{-3}$.

Would you like further clarification or details on any part of the solution?

Here are 5 follow-up questions:

1. How does the sum formula for a geometric sequence change if the common ratio is greater than 1?
2. What happens to the geometric sequence when the common ratio is negative?
3. Can the sum of an infinite geometric series exist when the common ratio is -3? Why or why not?
4. How do you derive the formula for the sum of a geometric series?
5. What would happen to the sum if the first term of the sequence were a different value?

Tip: In geometric sequences, if the absolute value of the common ratio is greater than 1, the terms grow quickly, leading to large sums or diverging series.