The fourth term of the arithmetic - geometric progression 6, 8, 8, ………. Is

Options

A. 8
B. 12
C. `(32)/(3)`
D. `(64)/(9)`

Answer

To find the fourth term of the arithmetic-geometric progression (AGP) given by the terms 6, 8, 8, ..., we can follow these steps:

### Step 1: Identify the first term and the second term
The first term \( a \) is 6 and the second term \( T_2 \) is 8.

### Step 2: Write the general term of the AGP
The general term of an AGP can be expressed as:
\[ T_n = a + (n - 1)d \cdot r^{n - 1} \]
where:
- \( a \) is the first term,
- \( d \) is the common difference of the arithmetic sequence,
- \( r \) is the common ratio of the geometric sequence.

### Step 3: Set up equations using known terms
Using the second term:
\[ T_2 = a + d \cdot r = 8 \]
Substituting \( a = 6 \):
\[ 6 + d \cdot r = 8 \implies d \cdot r = 2 \quad (1) \]

Using the third term:
\[ T_3 = a + 2d \cdot r^2 = 8 \]
Substituting \( a = 6 \):
\[ 6 + 2d \cdot r^2 = 8 \implies 2d \cdot r^2 = 2 \implies d \cdot r^2 = 1 \quad (2) \]

### Step 4: Solve the equations
From equation (1):
\[ d \cdot r = 2 \quad (1) \]
From equation (2):
\[ d \cdot r^2 = 1 \quad (2) \]

Now, divide equation (2) by equation (1):
\[ \frac{d \cdot r^2}{d \cdot r} = \frac{1}{2} \]
This simplifies to:
\[ r = \frac{1}{2} \]

### Step 5: Substitute \( r \) back to find \( d \)
Substituting \( r = \frac{1}{2} \) back into equation (1):
\[ d \cdot \frac{1}{2} = 2 \implies d = 4 \]

### Step 6: Find the fourth term \( T_4 \)
Now we can find the fourth term using:
\[ T_4 = a + 3d \cdot r^3 \]
Substituting \( a = 6 \), \( d = 4 \), and \( r = \frac{1}{2} \):
\[ T_4 = 6 + 3 \cdot 4 \cdot \left(\frac{1}{2}\right)^3 \]
Calculating \( \left(\frac{1}{2}\right)^3 = \frac{1}{8} \):
\[ T_4 = 6 + 3 \cdot 4 \cdot \frac{1}{8} = 6 + \frac{12}{8} = 6 + \frac{3}{2} = 6 + 1.5 = 7.5 \]

### Final Answer
The fourth term of the arithmetic-geometric progression is \( 7.5 \).

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