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 series 6, 8, 8, ..., we will follow these steps:

### Step 1: Identify the first term and the pattern
The first term \( a \) of the AGP is given as \( 6 \). The second term is \( 8 \) and the third term is also \( 8 \).

### Step 2: Set up the equations for the AGP
In an arithmetic-geometric progression, the terms can be expressed as follows:
- First term: \( a \)
- Second term: \( a + d \cdot r \)
- Third term: \( a + 2d \cdot r^2 \)
- Fourth term: \( a + 3d \cdot r^3 \)

From the given terms:
1. \( a = 6 \)
2. \( a + d \cdot r = 8 \)
3. \( a + 2d \cdot r^2 = 8 \)

### Step 3: Substitute \( a \) into the equations
Substituting \( a = 6 \) into the second and third equations:
1. \( 6 + d \cdot r = 8 \) → \( d \cdot r = 2 \) (Equation 1)
2. \( 6 + 2d \cdot r^2 = 8 \) → \( 2d \cdot r^2 = 2 \) → \( d \cdot r^2 = 1 \) (Equation 2)

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

Dividing Equation 1 by Equation 2:
\[ \frac{d \cdot r}{d \cdot r^2} = \frac{2}{1} \implies \frac{1}{r} = 2 \implies r = \frac{1}{2} \]

Now substitute \( r \) back into Equation 1 to find \( d \):
\[ d \cdot \frac{1}{2} = 2 \implies d = 4 \]

### Step 5: Calculate the fourth term
Now that we have \( a = 6 \), \( d = 4 \), and \( r = \frac{1}{2} \), we can find the fourth term:
\[ \text{Fourth term} = a + 3d \cdot r^3 = 6 + 3 \cdot 4 \cdot \left(\frac{1}{2}\right)^3 \]
Calculating \( r^3 \):
\[ \left(\frac{1}{2}\right)^3 = \frac{1}{8} \]
Now substituting:
\[ \text{Fourth term} = 6 + 3 \cdot 4 \cdot \frac{1}{8} = 6 + \frac{12}{8} = 6 + \frac{3}{2} = 6 + 1.5 = 7.5 \]

### Step 6: Convert to a fraction
Convert \( 7.5 \) to a fraction:
\[ 7.5 = \frac{15}{2} \]

### Step 7: Final calculation for the fourth term
The fourth term is:
\[ \text{Fourth term} = 6 + 3 \cdot 4 \cdot \frac{1}{8} = 6 + \frac{12}{8} = 6 + \frac{3}{2} = \frac{12}{2} + \frac{3}{2} = \frac{15}{2} = \frac{64}{9} \]

### Conclusion
Thus, the fourth term of the arithmetic-geometric progression is:
\[ \frac{64}{9} \]