If `a,a^(2)+2,a^(3)+10` be three consecutive terms of G.P., then the fourth term is

Options

A. 0
B. 6
C. `(729)/(16)`
D. 54

Answer

To find the fourth term of the geometric progression (G.P.) given the first three terms \( a, a^2 + 2, a^3 + 10 \), we will follow these steps:

### Step 1: Identify the terms
Let:
- First term \( T_1 = a \)
- Second term \( T_2 = a^2 + 2 \)
- Third term \( T_3 = a^3 + 10 \)

### Step 2: Use the property of G.P.
In a G.P., the square of the middle term is equal to the product of the other two terms. Therefore, we can write:
\[ (T_2)^2 = T_1 \cdot T_3 \]
Substituting the terms:
\[ (a^2 + 2)^2 = a \cdot (a^3 + 10) \]

### Step 3: Expand both sides
Expanding the left-hand side:
\[ (a^2 + 2)^2 = a^4 + 4a^2 + 4 \]
Expanding the right-hand side:
\[ a \cdot (a^3 + 10) = a^4 + 10a \]

### Step 4: Set up the equation
Now we set the two expansions equal to each other:
\[ a^4 + 4a^2 + 4 = a^4 + 10a \]

### Step 5: Simplify the equation
Subtract \( a^4 \) from both sides:
\[ 4a^2 + 4 = 10a \]
Rearranging gives:
\[ 4a^2 - 10a + 4 = 0 \]

### Step 6: Factor the quadratic equation
To factor the quadratic equation, we can rewrite it:
\[ 2a^2 - 5a + 2 = 0 \]
Now we can factor it:
\[ (2a - 1)(a - 2) = 0 \]

### Step 7: Solve for \( a \)
Setting each factor to zero gives:
1. \( 2a - 1 = 0 \) → \( a = \frac{1}{2} \)
2. \( a - 2 = 0 \) → \( a = 2 \)

### Step 8: Find the common ratio \( r \)
Now we can find the common ratio \( r \) for both values of \( a \).

**For \( a = \frac{1}{2} \):**
\[ T_2 = a^2 + 2 = \left(\frac{1}{2}\right)^2 + 2 = \frac{1}{4} + 2 = \frac{9}{4} \]
\[ r = \frac{T_2}{T_1} = \frac{\frac{9}{4}}{\frac{1}{2}} = \frac{9}{4} \cdot \frac{2}{1} = \frac{9}{2} \]

**For \( a = 2 \):**
\[ T_2 = a^2 + 2 = 2^2 + 2 = 4 + 2 = 6 \]
\[ r = \frac{T_2}{T_1} = \frac{6}{2} = 3 \]

### Step 9: Calculate the fourth term
The fourth term \( T_4 \) in a G.P. can be calculated using the formula:
\[ T_4 = T_1 \cdot r^3 \]

**For \( a = \frac{1}{2} \):**
\[ T_4 = \frac{1}{2} \cdot \left(\frac{9}{2}\right)^3 = \frac{1}{2} \cdot \frac{729}{8} = \frac{729}{16} \]

**For \( a = 2 \):**
\[ T_4 = 2 \cdot 3^3 = 2 \cdot 27 = 54 \]

### Conclusion
The fourth term can be either \( \frac{729}{16} \) or \( 54 \).