Find the sum to 8 terms of
`3 + 6 + 12 + ...`

Answer

To find the sum of the first 8 terms of the series \(3 + 6 + 12 + \ldots\), we first need to identify the type of sequence we are dealing with.

### Step-by-Step Solution:

1. **Identify the Sequence Type**:
The given series is \(3, 6, 12, \ldots\). To check if it is a geometric progression (GP), we can calculate the ratio of consecutive terms:
- Second term divided by the first term:
\[ \frac{6}{3} = 2 \]
- Third term divided by the second term:
\[ \frac{12}{6} = 2 \]
Since the ratio is constant, this is a geometric progression.

2. **Identify the First Term and Common Ratio**:
- The first term \(A\) is \(3\).
- The common ratio \(R\) is \(2\).

3. **Use the Formula for the Sum of the First n Terms of a GP**:
The formula for the sum of the first \(n\) terms of a geometric progression is:
\[ S_n = A \frac{R^n - 1}{R - 1} \]
Here, we need to find \(S_8\), so \(n = 8\).

4. **Substitute the Values into the Formula**:
Now substituting \(A = 3\), \(R = 2\), and \(n = 8\) into the formula:
\[ S_8 = 3 \frac{2^8 - 1}{2 - 1} \]

5. **Calculate \(2^8\)**:
\[ 2^8 = 256 \]

6. **Calculate \(S_8\)**:
Now substituting \(2^8\) back into the equation:
\[ S_8 = 3 \frac{256 - 1}{1} = 3 \times 255 \]
\[ S_8 = 765 \]

Thus, the sum of the first 8 terms of the series is \(765\).

### Final Answer:
The sum to 8 terms of the series \(3 + 6 + 12 + \ldots\) is \(765\).