In a G.P of positive terms if any term is equal to the sum of the next two terms, then the common ratio of the G.P is

Options

A. `sin18^(@)`
B. `2cos18^(@)`
C. `cos18^(@)`
D. `2sin18^(@)`

Answer

To solve the problem, we need to find the common ratio \( R \) of a geometric progression (G.P.) where any term is equal to the sum of the next two terms.

### Step-by-step Solution:

1. **Define the terms of the G.P.:**
Let the first term of the G.P. be \( A \) and the common ratio be \( R \). The \( n \)-th term of the G.P. can be expressed as:
\[ A_n = A R^{n-1} \]

2. **Set up the equation based on the problem statement:**
According to the problem, any term \( A_n \) is equal to the sum of the next two terms \( A_{n+1} \) and \( A_{n+2} \). This gives us the equation:
\[ A R^{n-1} = A R^n + A R^{n+1} \]

3. **Simplify the equation:**
We can factor out \( A \) from both sides (assuming \( A \neq 0 \)):
\[ R^{n-1} = R^n + R^{n+1} \]
Dividing through by \( R^{n-1} \) (assuming \( R \neq 0 \)):
\[ 1 = R + R^2 \]

4. **Rearrange to form a quadratic equation:**
Rearranging the equation gives us:
\[ R^2 + R - 1 = 0 \]

5. **Solve the quadratic equation using the quadratic formula:**
The quadratic formula is given by:
\[ R = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 1, b = 1, c = -1 \):
\[ R = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-1)}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 + 4}}{2} = \frac{-1 \pm \sqrt{5}}{2} \]

6. **Determine the valid solution:**
Since the common ratio \( R \) must be positive, we take the positive root:
\[ R = \frac{-1 + \sqrt{5}}{2} \]

7. **Relate to trigonometric values:**
We can express this result in terms of trigonometric functions. Notably, we know that:
\[ \sin(18^\circ) = \frac{\sqrt{5} - 1}{4} \]
Therefore, we can rewrite \( R \) as:
\[ R = 2 \sin(18^\circ) \]

### Final Answer:
The common ratio \( R \) of the G.P. is:
\[ R = 2 \sin(18^\circ) \]