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I can help extract the math problems from your image and solve them. Let me first transcribe the relevant parts of the image.

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1. **Problem 6:**  
_In a GP where terms are all positive, the third term exceeds the first term by 32, while the fifth term exceeds the first term by 320. Calculate the first term._

2. **Problem 7:**  
_Find the geometric means of the following:_  
    (a) \( 3, 12 \)  
    (b) \( 4, 36 \)  
    (c) \( 5, 20, 45 \)

3. **Problem 8:**  
_Find the indicated sums of these GPs:_  
    (i) \( 5, 10, 20, \dots \) find \( S_6 \) and \( S_8 \).  
    (ii) \( 3, -6, 12, \dots \) find \( S_5 \) and \( S_6 \).

4. **Problem 9:**  
_Given \( a + a r + a r^2 + a r^3 = 40 \), then the first positive term of a GP, calculate the sum of the first 6 terms of the GP._

5. **Problem 11:**  
_The sum of the first 3 terms of a GP is 117, and the sum of the next 3 terms is 36. Calculate the third term._

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I will begin solving these one by one. Let's start with **Problem 6**.

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### Problem 6:

Given that:
- The third term exceeds the first term by 32, i.e., \( T_3 = a + 32 \).
- The fifth term exceeds the first term by 320, i.e., \( T_5 = a + 320 \).

The general formula for the \(n\)-th term of a geometric progression (GP) is:
\[
T_n = a \cdot r^{n-1}
\]

For the third term:
\[
T_3 = a \cdot r^2 = a + 32
\]

For the fifth term:
\[
T_5 = a \cdot r^4 = a + 320
\]

Now, I will solve this system of equations to find \(a\) (the first term) and \(r\) (the common ratio). Let's compute.The solutions for the first term \( a \) and the common ratio \( r \) are:
- \( a = 4 \), \( r = -3 \)
- \( a = 4 \), \( r = 3 \)

Since the terms of the GP are stated to be positive, we discard the solution with \( r = -3 \). Therefore, the first term is \( a = 4 \) and the common ratio is \( r = 3 \).

Would you like to proceed with solving the next problem, or do you have any questions on this solution?

Here are some relative questions to expand on this problem:
1. What is the formula for the \(n\)-th term of a geometric progression?
2. How does the common ratio affect the growth of a geometric progression?
3. What is the sum of the first 5 terms of this GP?
4. What happens to the sequence if the common ratio is negative?
5. How do we find the sum of infinite terms in a geometric progression?

**Tip:** In a GP, if the common ratio is greater than 1, the terms grow rapidly.

In a GP where terms are all positive, the third term exceeds the first term by 32 while the fifth term exceeds the first term by 320. Calculate the first term.

Learn how to solve a geometric progression (GP) problem where the third and fifth terms exceed the first term by specific amounts. This solution demonstrates how to apply the formula for the nth term of a GP and solve for the first term and common ratio.

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