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Let's analyze and solve the given problems from the image:

### 1. For each of the following GPs, find the common ratio \(r\).
   A geometric progression (GP) has a constant ratio between consecutive terms. The common ratio \(r\) can be calculated by dividing the second term by the first term.

   **(a)** \(3, 9, 27, \dots\)  
   \(r = \frac{9}{3} = 3\)

   **(b)** \(10, 5, \frac{5}{2}, \dots\)  
   \(r = \frac{5}{10} = \frac{1}{2}\)

   **(c)** \(3\sqrt{3}, 3\sqrt{6}, 6\sqrt{3}, \dots\)  
   \(r = \frac{3\sqrt{6}}{3\sqrt{3}} = \frac{\sqrt{6}}{\sqrt{3}} = \sqrt{2}\)

   **(d)** \( x + x^2, x + x^3, x + x^4, \dots\)  
   \(r = \frac{x + x^3}{x + x^2} = \frac{x(1 + x^2)}{x(1 + x)} = \frac{1 + x^2}{1 + x}\)

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### 2. Find the indicated terms of these GPs.
   The general term for a geometric progression is \( T_n = ar^{n-1} \), where \(a\) is the first term and \(r\) is the common ratio.

   **(a)** \(2, 10, 50, \dots \)  
   - First term \(a = 2\)
   - Common ratio \(r = \frac{10}{2} = 5\)
   
   Now, let's find the 7th term:  
   \(T_7 = ar^{7-1} = 2 \times 5^6 = 2 \times 15625 = 31250\)

   **(b)** \( \frac{1}{2}, 2, 8, \dots\)  
   - First term \(a = \frac{1}{2}\)
   - Common ratio \(r = \frac{2}{\frac{1}{2}} = 4\)
   
   Find the 8th term:  
   \(T_8 = ar^{8-1} = \frac{1}{2} \times 4^7 = \frac{1}{2} \times 16384 = 8192\)

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### 3. Find the fifth term of the GP \( \frac{1}{27}, \frac{9}{81}, 3, \dots \). How many terms are there?
   First, find the common ratio:
   \(r = \frac{\frac{9}{81}}{\frac{1}{27}} = \frac{9}{81} \times \frac{27}{1} = \frac{9 \times 27}{81} = 3\)

   Now, find the 5th term:  
   \(T_5 = ar^{5-1} = \frac{1}{27} \times 3^4 = \frac{1}{27} \times 81 = 3\)

   To find the total number of terms, note that the last term is 81, which is written as:
   \[
   T_n = ar^{n-1} = \frac{1}{27} \times 3^{n-1} = 81
   \]
   Solve for \(n\):  
   \[
   \frac{1}{27} \times 3^{n-1} = 81
   \]
   \[
   3^{n-1} = 81 \times 27 = 2187 = 3^7
   \]
   \[
   n - 1 = 7 \quad \Rightarrow \quad n = 8
   \]

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### 4. In a GP, the fourth term is 18 and the seventh term is 486. Find the second term.
   Let the first term be \(a\) and the common ratio be \(r\).

   - Fourth term: \( T_4 = ar^3 = 18 \)
   - Seventh term: \( T_7 = ar^6 = 486 \)

   Dividing the second equation by the first to eliminate \(a\):  
   \[
   \frac{

For each of the following GPs, find the common ratio and solve other related questions about geometric progressions.

This detailed solution covers the process of finding the common ratio in geometric progressions (GP), calculating the Nth term, and solving for the sum of terms. Topics include common ratio formulas and the Nth term equation, ideal for students in Grades 10-12.

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