If `a,2a+2,3a+3` are in GP, then what is the fourth term of the GP?

Options

A. `-13.5`
B. `13.5`
C. -27
D. 27

Explanation

To solve the problem where the terms \( a, 2a + 2, 3a + 3 \) are in geometric progression (GP), we will follow these steps:

### Step 1: Understand the condition for GP
For three numbers \( a, b, c \) to be in GP, the condition is:
\[ b^2 = ac \]
In our case, \( a = a \), \( b = 2a + 2 \), and \( c = 3a + 3 \).

### Step 2: Set up the equation
Using the condition for GP:
\[ (2a + 2)^2 = a(3a + 3) \]

### Step 3: Expand both sides
Expanding the left side:
\[ (2a + 2)^2 = 4a^2 + 8a + 4 \]
Expanding the right side:
\[ a(3a + 3) = 3a^2 + 3a \]

### Step 4: Set the equation
Now we equate both sides:
\[ 4a^2 + 8a + 4 = 3a^2 + 3a \]

### Step 5: Rearrange the equation
Rearranging gives:
\[ 4a^2 - 3a^2 + 8a - 3a + 4 = 0 \]
This simplifies to:
\[ a^2 + 5a + 4 = 0 \]

### Step 6: Factor the quadratic equation
Now we factor the quadratic:
\[ (a + 4)(a + 1) = 0 \]

### Step 7: Solve for \( a \)
Setting each factor to zero gives:
\[ a + 4 = 0 \quad \Rightarrow \quad a = -4 \]
\[ a + 1 = 0 \quad \Rightarrow \quad a = -1 \]

### Step 8: Find the fourth term of the GP
Now we need to find the fourth term of the GP for both values of \( a \).

1. **For \( a = -4 \)**:
- First term: \( -4 \)
- Second term: \( 2(-4) + 2 = -8 + 2 = -6 \)
- Third term: \( 3(-4) + 3 = -12 + 3 = -9 \)
- The common ratio \( r \) can be calculated as:
\[ r = \frac{-6}{-4} = \frac{3}{2} \]
- Fourth term:
\[ \text{Fourth term} = -9 \times \frac{3}{2} = -\frac{27}{2} \]

2. **For \( a = -1 \)**:
- First term: \( -1 \)
- Second term: \( 2(-1) + 2 = -2 + 2 = 0 \)
- Third term: \( 3(-1) + 3 = -3 + 3 = 0 \)
- Since the second and third terms are both 0, the GP does not progress further.

### Conclusion
The fourth term of the GP, when \( a = -4 \), is:
\[ \text{Fourth term} = -\frac{27}{2} \]