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.