The sum of the infinite Arithmetico -Geometric progression3,4,4,… is _________.

Explanation

To find the sum of the infinite Arithmetico-Geometric progression (AGP) given as 3, 4, 4, ..., we can follow these steps:

### Step 1: Identify the first term (A), common difference (D), and common ratio (r)
The first term \( A \) is given as 3. The second term is 4, and the third term is also 4.

We can express the terms of the AGP as follows:
- First term: \( A = 3 \)
- Second term: \( A + D \cdot r = 4 \)
- Third term: \( A + 2D \cdot r^2 = 4 \)

### Step 2: Set up equations based on the terms
From the second term:
\[ A + D \cdot r = 4 \]
Substituting \( A = 3 \):
\[ 3 + D \cdot r = 4 \]
Thus,
\[ D \cdot r = 4 - 3 = 1 \]
So, we have:
\[ D \cdot r = 1 \quad \text{(1)} \]

From the third term:
\[ A + 2D \cdot r^2 = 4 \]
Substituting \( A = 3 \):
\[ 3 + 2D \cdot r^2 = 4 \]
Thus,
\[ 2D \cdot r^2 = 4 - 3 = 1 \]
So, we have:
\[ D \cdot r^2 = \frac{1}{2} \quad \text{(2)} \]

### Step 3: Solve for D and r
From equation (1):
\[ D = \frac{1}{r} \]

Substituting \( D \) in equation (2):
\[ \frac{1}{r} \cdot r^2 = \frac{1}{2} \]
This simplifies to:
\[ r = \frac{1}{2} \]

Now substituting \( r \) back into equation (1) to find \( D \):
\[ D \cdot \frac{1}{2} = 1 \]
Thus,
\[ D = 2 \]

### Step 4: Calculate the sum of the infinite AGP
The formula for the sum \( S \) of an infinite AGP is:
\[ S = \frac{A}{1 - r} + \frac{D \cdot r}{(1 - r)^2} \]
Substituting \( A = 3 \), \( D = 2 \), and \( r = \frac{1}{2} \):
\[ S = \frac{3}{1 - \frac{1}{2}} + \frac{2 \cdot \frac{1}{2}}{\left(1 - \frac{1}{2}\right)^2} \]
Calculating each term:
\[ S = \frac{3}{\frac{1}{2}} + \frac{1}{\left(\frac{1}{2}\right)^2} \]
\[ S = 6 + 4 = 10 \]

### Final Answer
The sum of the infinite Arithmetico-Geometric progression is **10**.

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