Find the sum of 7 terms of the series 3√2, 12, 12√2

We are given a series: \( 3\sqrt{2}, 12, 12\sqrt{2}, \dots \). This appears to be a geometric progression (GP) because each term is multiplied by a common ratio.

### Step 1: Verify the common ratio
The first term of the series is \( a_1 = 3\sqrt{2} \). To find the common ratio \( r \), we divide the second term by the first term:

\[
r = \frac{12}{3\sqrt{2}} = \frac{12}{3} \times \frac{1}{\sqrt{2}} = 4 \times \frac{\sqrt{2}}{2} = 2\sqrt{2}
\]

So, the common ratio is \( r = 2\sqrt{2} \).

### Step 2: Use the sum formula for the first 7 terms of a GP
The formula for the sum of the first \( n \) terms of a geometric progression is:

\[
S_n = a \frac{1 - r^n}{1 - r}, \text{ for } r \neq 1
\]

Here:
- \( a = 3\sqrt{2} \)
- \( r = 2\sqrt{2} \)
- \( n = 7 \)

### Step 3: Compute the sum
We will substitute these values into the formula:

\[
S_7 = 3\sqrt{2} \frac{1 - (2\sqrt{2})^7}{1 - 2\sqrt{2}}
\]

First, let's simplify \( (2\sqrt{2})^7 \):

\[
(2\sqrt{2})^7 = 2^7 \times (\sqrt{2})^7 = 128 \times 2^{3.5} = 128 \times 11.3137 \approx 1448.15
\]

Now substitute this back:

\[
S_7 = 3\sqrt{2} \frac{1 - 1448.15}{1 - 2\sqrt{2}} = 3\sqrt{2} \frac{-1447.15}{1 - 2.828}
\]

Since the denominator is negative:

\[
S_7 = 3\sqrt{2} \frac{-1447.15}{-1.828} \approx 3\sqrt{2} \times 791.73
\]

Now multiply:

\[
S_7 \approx 3 \times 1.414 \times 791.73 \approx 3358.73
\]

Thus, the sum of the first 7 terms is approximately \( 3358.73 \).

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Would you like more details or clarifications? 

Here are five related questions you might explore:
1. How do you find the sum of an infinite geometric series?
2. What happens when the common ratio \( r \) is between -1 and 1?
3. How do geometric progressions apply in real-life situations?
4. Can you find the 8th term of this GP?
5. How does the sum of an arithmetic progression differ from that of a geometric one?

**Tip:** Always check the common ratio carefully when determining the type of sequence.

Learn how to find the sum of the first 7 terms of the geometric progression series 3√2, 12, 12√2. This solution explains the step-by-step process, including calculating the common ratio and applying the geometric series sum formula. Suitable for students in Grades 9-12.

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