if `b_(1),b_(2),b_(3)(b_(1)gt0)` are three successive terms of a G.P. with common ratio `r`, the value of for which the inequality `b_(3)gt4b_(2)-3b_(1)`, holds is given by

Options

A. `rgt3`
B. `0ltrlt1`
C. `r=3.5`
D. `r=5.2`

Explanation

To solve the inequality \( b_3 > 4b_2 - 3b_1 \) given that \( b_1, b_2, b_3 \) are three successive terms of a geometric progression (G.P.) with common ratio \( r \), we can follow these steps:

### Step 1: Express the terms of the G.P.
Let:
- \( b_1 = a \)
- \( b_2 = ar \)
- \( b_3 = ar^2 \)

### Step 2: Substitute the terms into the inequality
We need to substitute \( b_1, b_2, \) and \( b_3 \) into the inequality:
\[ b_3 > 4b_2 - 3b_1 \]
Substituting the expressions we have:
\[ ar^2 > 4(ar) - 3a \]

### Step 3: Simplify the inequality
Now, simplify the right side:
\[ ar^2 > 4ar - 3a \]
We can factor out \( a \) from the right side (since \( a > 0 \)):
\[ ar^2 > a(4r - 3) \]
Dividing both sides by \( a \) (which is positive):
\[ r^2 > 4r - 3 \]

### Step 4: Rearrange the inequality
Rearranging gives:
\[ r^2 - 4r + 3 > 0 \]

### Step 5: Factor the quadratic expression
Now, we can factor the quadratic:
\[ (r - 1)(r - 3) > 0 \]

### Step 6: Determine the intervals
To solve the inequality \( (r - 1)(r - 3) > 0 \), we find the critical points where the expression equals zero:
- \( r = 1 \)
- \( r = 3 \)

Now, we test the intervals:
1. For \( r < 1 \): Choose \( r = 0 \) → \( (0 - 1)(0 - 3) = 3 > 0 \) (True)
2. For \( 1 < r < 3 \): Choose \( r = 2 \) → \( (2 - 1)(2 - 3) = -1 < 0 \) (False)
3. For \( r > 3 \): Choose \( r = 4 \) → \( (4 - 1)(4 - 3) = 3 > 0 \) (True)

### Step 7: Conclusion
The solution to the inequality is:
\[ r < 1 \quad \text{or} \quad r > 3 \]