Let $S_n$ be the sum of the first $n$ terms of a geometric sequence $\{a_n\}$. Given $3S_3 = a_4 - 3$ and $3S_2 = a_3 - 3$, find the common ratio $q$.

1. Problem Description

Let $S_n$ be the sum of the first $n$ terms of a geometric sequence $\{a_n\}$. Given $3S_3 = a_4 - 3$ and $3S_2 = a_3 - 3$, find the common ratio $q$.

2. Solution Steps

We are given that $3S_3 = a_4 - 3$ and $3S_2 = a_3 - 3$.

Since $S_n$ is the sum of the first $n$ terms of a geometric sequence, we have

$S_n = a_1 + a_2 + \dots + a_n$.

Also, $a_n = a_1 q^{n-1}$, where $a_1$ is the first term and $q$ is the common ratio.

We can express $S_2$ and $S_3$ as:

$S_2 = a_1 + a_2 = a_1 + a_1q$

$S_3 = a_1 + a_2 + a_3 = a_1 + a_1q + a_1q^2$

From $3S_3 = a_4 - 3$, we have $3(a_1 + a_1q + a_1q^2) = a_1q^3 - 3$, which can be written as:

$3a_1 + 3a_1q + 3a_1q^2 = a_1q^3 - 3$

$a_1q^3 - 3a_1q^2 - 3a_1q - 3a_1 = 3$ (1)

From $3S_2 = a_3 - 3$, we have $3(a_1 + a_1q) = a_1q^2 - 3$, which can be written as:

$3a_1 + 3a_1q = a_1q^2 - 3$

$a_1q^2 - 3a_1q - 3a_1 = 3$ (2)

Comparing (1) and (2), we have:

$a_1q^3 - 3a_1q^2 - 3a_1q - 3a_1 = a_1q^2 - 3a_1q - 3a_1$

$a_1q^3 - 4a_1q^2 = 0$

$a_1q^2(q - 4) = 0$

Since $a_1 \neq 0$ and $q \neq 0$ (otherwise the sequence is trivial), we must have $q - 4 = 0$, so $q = 4$.

3. Final Answer

The common ratio $q = 4$.