If `a_(1),a_(2),a_(3)(a_(1)gt0)` are three successive terms of a GP with common ratio r, the value of r for which `a_(3)gt4a_(2)-3a_(1)` holds is given by

Options

A. `1ltrlt3`
B. `-3ltrlt-1`
C. `rlt1` or `rgt3`
D. None of these

Answer

To solve the problem, we need to find the value of the common ratio \( r \) for which the inequality \( a_3 > 4a_2 - 3a_1 \) holds, given that \( a_1, a_2, a_3 \) are three successive terms of a geometric progression (GP).

### Step-by-Step Solution:

1. **Define the terms of the GP**:
Let \( a_1 = a \), \( a_2 = ar \), and \( a_3 = ar^2 \), where \( a > 0 \) and \( r \) is the common ratio.

2. **Set up the inequality**:
We need to analyze the inequality:
\[ a_3 > 4a_2 - 3a_1 \]
Substituting the terms of the GP into the inequality, we have:
\[ ar^2 > 4(ar) - 3a \]

3. **Simplify the inequality**:
Rearranging the terms gives:
\[ ar^2 > 4ar - 3a \]
Dividing through by \( a \) (since \( a > 0 \)):
\[ r^2 > 4r - 3 \]

4. **Rearrange into standard quadratic form**:
Rearranging the above inequality results in:
\[ r^2 - 4r + 3 > 0 \]

5. **Factor the quadratic**:
The quadratic \( r^2 - 4r + 3 \) can be factored as:
\[ (r - 1)(r - 3) > 0 \]

6. **Determine the intervals**:
To solve the inequality \( (r - 1)(r - 3) > 0 \), we find the critical points, which are \( r = 1 \) and \( r = 3 \). We can use the test point method or the wavy curve method to determine the sign of the expression in the intervals defined by these points:
- For \( r < 1 \): Choose \( r = 0 \) → \( (0 - 1)(0 - 3) = 3 > 0 \) (positive)
- For \( 1 < r < 3 \): Choose \( r = 2 \) → \( (2 - 1)(2 - 3) = -1 < 0 \) (negative)
- For \( r > 3 \): Choose \( r = 4 \) → \( (4 - 1)(4 - 3) = 3 > 0 \) (positive)

7. **Combine the intervals**:
The solution to the inequality \( (r - 1)(r - 3) > 0 \) is:
\[ r < 1 \quad \text{or} \quad r > 3 \]

### Final Answer:
Thus, the values of \( r \) for which the inequality holds are:
\[ r \in (-\infty, 1) \cup (3, \infty) \]