If `In (2x^2 - 5), In (x^2 - 1)` and `In(x^2 - 3)` are the first three terms of an arithmetic progression, then its fourth term is

Options

A. In 8 - In 3
B. In 3 - In 8
C. In 24
D. 2 In 6

Answer

To solve the problem, we need to determine the fourth term of an arithmetic progression (AP) given the first three terms as \( \ln(2x^2 - 5) \), \( \ln(x^2 - 1) \), and \( \ln(x^2 - 3) \).

### Step-by-Step Solution:

1. **Identify the Terms in AP**:
The terms are:
- \( a = \ln(2x^2 - 5) \)
- \( b = \ln(x^2 - 1) \)
- \( c = \ln(x^2 - 3) \)

2. **Use the AP Condition**:
For three terms to be in AP, the following condition must hold:
\[ 2b = a + c \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ 2\ln(x^2 - 1) = \ln(2x^2 - 5) + \ln(x^2 - 3) \]

3. **Apply Logarithmic Properties**:
Using the property \( \ln(a) + \ln(b) = \ln(ab) \), we can rewrite the right side:
\[ 2\ln(x^2 - 1) = \ln((2x^2 - 5)(x^2 - 3)) \]
Now, we can express \( 2\ln(x^2 - 1) \) as:
\[ \ln((x^2 - 1)^2) = \ln((2x^2 - 5)(x^2 - 3)) \]

4. **Equate the Arguments**:
Since the logarithms are equal, we can equate the arguments:
\[ (x^2 - 1)^2 = (2x^2 - 5)(x^2 - 3) \]

5. **Expand Both Sides**:
- Left side:
\[ (x^2 - 1)^2 = x^4 - 2x^2 + 1 \]
- Right side:
\[ (2x^2 - 5)(x^2 - 3) = 2x^4 - 6x^2 - 5x^2 + 15 = 2x^4 - 11x^2 + 15 \]

6. **Set Up the Equation**:
Now we have:
\[ x^4 - 2x^2 + 1 = 2x^4 - 11x^2 + 15 \]

7. **Rearrange the Equation**:
Rearranging gives:
\[ 0 = 2x^4 - x^4 - 11x^2 + 2x^2 + 15 - 1 \]
Simplifying:
\[ 0 = x^4 - 9x^2 + 14 \]

8. **Factor the Quadratic**:
This can be factored as:
\[ (x^2 - 7)(x^2 - 2) = 0 \]
Thus, \( x^2 = 7 \) or \( x^2 = 2 \).

9. **Find the Fourth Term**:
The fourth term \( a_4 \) in an AP is given by:
\[ a_4 = a + 3d \]
where \( d = b - a \).

First, calculate \( d \):
\[ d = \ln(x^2 - 1) - \ln(2x^2 - 5) = \ln\left(\frac{x^2 - 1}{2x^2 - 5}\right) \]

Now, substituting \( a \) and \( d \):
\[ a_4 = \ln(2x^2 - 5) + 3\ln\left(\frac{x^2 - 1}{2x^2 - 5}\right) \]
This simplifies to:
\[ a_4 = \ln(2x^2 - 5) + \ln\left(\frac{(x^2 - 1)^3}{(2x^2 - 5)^3}\right) \]
\[ a_4 = \ln\left((2x^2 - 5) \cdot \frac{(x^2 - 1)^3}{(2x^2 - 5)^3}\right) = \ln\left(\frac{(x^2 - 1)^3}{(2x^2 - 5)^2}\right) \]

10. **Evaluate for Each Case**:
- For \( x^2 = 2 \):
\[ a_4 = \ln\left(\frac{(2 - 1)^3}{(4 - 5)^2}\right) = \ln\left(\frac{1^3}{(-1)^2}\right) = \ln(1) = 0 \]
- For \( x^2 = 7 \):
\[ a_4 = \ln\left(\frac{(7 - 1)^3}{(14 - 5)^2}\right) = \ln\left(\frac{6^3}{9^2}\right) = \ln\left(\frac{216}{81}\right) = \ln\left(\frac{8}{3}\right) \]

### Final Answer:
Thus, the fourth term can be either \( 0 \) or \( \ln\left(\frac{8}{3}\right) \).