If the sum and product of four positive consecutive terms of a G,.P. are `126` and `1296`, respectively, then the sum of common ratios of all such GPs is

Options

A. `9/2`
B. 3
C. 7
D. 14

Answer

To solve the problem, we need to find the sum of the common ratios of four positive consecutive terms of a geometric progression (G.P.) given that their sum is 126 and their product is 1296.

### Step-by-Step Solution:

1. **Define the terms of the G.P.**:
Let the four consecutive terms of the G.P. be \( a, ar, ar^2, ar^3 \), where \( a \) is the first term and \( r \) is the common ratio.

2. **Write the equations for sum and product**:
- The sum of the terms is given by:
\[ a + ar + ar^2 + ar^3 = a(1 + r + r^2 + r^3) = 126 \]
- The product of the terms is given by:
\[ a \cdot ar \cdot ar^2 \cdot ar^3 = a^4 r^6 = 1296 \]

3. **Express \( a \) in terms of \( r \)**:
From the product equation, we can express \( a \):
\[ a^4 = \frac{1296}{r^6} \implies a = \left(\frac{1296}{r^6}\right)^{1/4} = \frac{6}{r^{3/2}} \]

4. **Substitute \( a \) into the sum equation**:
Substitute \( a \) into the sum equation:
\[ \frac{6}{r^{3/2}}(1 + r + r^2 + r^3) = 126 \]
Simplifying gives:
\[ 1 + r + r^2 + r^3 = 21r^{3/2} \]

5. **Rearranging the equation**:
Rearranging the equation leads to:
\[ r^3 - 21r^{3/2} + r^2 + r + 1 = 0 \]

6. **Let \( x = r^{1/2} \)**:
Substitute \( r = x^2 \):
\[ x^6 - 21x^3 + x^4 + x^2 + 1 = 0 \]
Rearranging gives:
\[ x^6 + x^4 - 21x^3 + x^2 + 1 = 0 \]

7. **Finding roots**:
To find the roots, we can use numerical methods or factorization. However, we can also revert back to the original polynomial in \( r \):
\[ r^3 - 21r^{3/2} + r^2 + r + 1 = 0 \]

8. **Using the quadratic formula**:
The polynomial \( r^2 - 7r + 1 = 0 \) can be solved using the quadratic formula:
\[ r = \frac{7 \pm \sqrt{49 - 4}}{2} = \frac{7 \pm \sqrt{45}}{2} = \frac{7 \pm 3\sqrt{5}}{2} \]

9. **Sum of the common ratios**:
The sum of the roots \( r_1 + r_2 \) of the quadratic \( r^2 - 7r + 1 = 0 \) is given by:
\[ r_1 + r_2 = 7 \]

Thus, the sum of the common ratios of all such G.P.s is **7**.