Let `a_(1),a_(2),a_(3)`, be three positive numbers which are `G.P.` with common ratio r. The inequality `a_(3) gt a_(2) + 2a_(1)` do not holds if r is equal to

Options

A. 2
B. 1.5
C. 0.5
D. 2.5

Answer

To solve the problem, we need to analyze the given inequality involving three positive numbers in geometric progression (G.P.) with a common ratio \( r \).

### Step-by-Step Solution:

1. **Define the Terms in G.P.**:
Let the three numbers in G.P. be:
\[ a_1 = a, \quad a_2 = ar, \quad a_3 = ar^2 \]
where \( a > 0 \) and \( r > 0 \).

2. **Set Up the Inequality**:
We need to analyze the inequality:
\[ a_3 > a_2 + 2a_1 \]
Substituting the values of \( a_1, a_2, \) and \( a_3 \):
\[ ar^2 > ar + 2a \]

3. **Rearrange the Inequality**:
We can rearrange the inequality to isolate terms involving \( r \):
\[ ar^2 - ar - 2a > 0 \]
Factoring out \( a \) (since \( a > 0 \)):
\[ a(r^2 - r - 2) > 0 \]
Since \( a > 0 \), we can focus on the quadratic:
\[ r^2 - r - 2 > 0 \]

4. **Factor the Quadratic**:
We can factor the quadratic expression:
\[ r^2 - r - 2 = (r - 2)(r + 1) \]
Thus, we need to solve:
\[ (r - 2)(r + 1) > 0 \]

5. **Determine the Critical Points**:
The critical points are \( r = 2 \) and \( r = -1 \). We will analyze the sign of the expression in the intervals determined by these points.

6. **Test the Intervals**:
- For \( r < -1 \): Choose \( r = -2 \):
\[ (-2 - 2)(-2 + 1) = (-4)(-1) = 4 > 0 \]
- For \( -1 < r < 2 \): Choose \( r = 0 \):
\[ (0 - 2)(0 + 1) = (-2)(1) = -2 < 0 \]
- For \( r > 2 \): Choose \( r = 3 \):
\[ (3 - 2)(3 + 1) = (1)(4) = 4 > 0 \]

7. **Conclusion**:
The inequality \( (r - 2)(r + 1) > 0 \) holds for:
- \( r < -1 \)
- \( r > 2 \)

Therefore, the inequality does not hold in the interval \( -1 < r < 2 \).

### Final Answer:
The inequality \( a_3 > a_2 + 2a_1 \) does not hold if \( r \) is in the interval \( (-1, 2) \).