If the `2^(nd),5^(th)` and `9^(th)` terms of a non-constant arithmetic progression are in geometric progession, then the common ratio of this geometric progression is

Options

A. 1
B. `(7)/(4)`
C. `(8)/(5)`
D. `(4)/(3)`

Answer

To solve the problem, we need to find the common ratio of the geometric progression formed by the 2nd, 5th, and 9th terms of a non-constant arithmetic progression (AP). Let's denote the first term of the AP as \( a \) and the common difference as \( d \).

### Step-by-Step Solution:

1. **Identify the Terms of the AP:**
- The \( n^{th} \) term of an AP is given by the formula:
\[ T_n = a + (n - 1)d \]
- Therefore, we can write:
- 2nd term \( T_2 = a + (2 - 1)d = a + d \)
- 5th term \( T_5 = a + (5 - 1)d = a + 4d \)
- 9th term \( T_9 = a + (9 - 1)d = a + 8d \)

2. **Set Up the Condition for GP:**
- We know that if the terms \( T_2, T_5, T_9 \) are in geometric progression, then:
\[ T_5^2 = T_2 \cdot T_9 \]
- Substituting the terms we found:
\[ (a + 4d)^2 = (a + d)(a + 8d) \]

3. **Expand Both Sides:**
- Expanding the left side:
\[ (a + 4d)^2 = a^2 + 8ad + 16d^2 \]
- Expanding the right side:
\[ (a + d)(a + 8d) = a^2 + 8ad + ad + 8d^2 = a^2 + 9ad + 8d^2 \]

4. **Set the Expanded Equations Equal:**
- Now we have:
\[ a^2 + 8ad + 16d^2 = a^2 + 9ad + 8d^2 \]

5. **Simplify the Equation:**
- Cancel \( a^2 \) from both sides:
\[ 8ad + 16d^2 = 9ad + 8d^2 \]
- Rearranging gives:
\[ 16d^2 - 8d^2 = 9ad - 8ad \]
\[ 8d^2 = ad \]

6. **Solve for \( d \):**
- If \( d \neq 0 \) (since the AP is non-constant), we can divide both sides by \( d \):
\[ 8d = a \quad \Rightarrow \quad a = 8d \]

7. **Find the Common Ratio \( r \):**
- The common ratio \( r \) of the GP can be found using:
\[ r = \frac{T_5}{T_2} = \frac{a + 4d}{a + d} \]
- Substitute \( a = 8d \):
\[ r = \frac{8d + 4d}{8d + d} = \frac{12d}{9d} = \frac{12}{9} = \frac{4}{3} \]

### Final Answer:
The common ratio of the geometric progression is \( \frac{4}{3} \).