The sum of an infinite geometric progression is 6, If the sum of the first two terms is `9//2`, then what is the first term?

Options

A. 1
B. `5//2`
C. 3 or `3//2`
D. 9 or 3

Explanation

To solve the problem step by step, we will use the formulas for the sum of an infinite geometric progression and the sum of the first two terms of a geometric progression.

### Step 1: Understand the Formulas
The sum \( S \) of an infinite geometric progression (GP) is given by:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio (with \( |r| < 1 \)).

The sum of the first two terms of a GP is:
\[ S_2 = a + ar = a(1 + r) \]

### Step 2: Set Up the Equations
From the problem, we know:
1. The sum of the infinite GP is 6:
\[ \frac{a}{1 - r} = 6 \quad \text{(1)} \]
2. The sum of the first two terms is \( \frac{9}{2} \):
\[ a(1 + r) = \frac{9}{2} \quad \text{(2)} \]

### Step 3: Solve Equation (1) for \( a \)
From equation (1):
\[ a = 6(1 - r) \quad \text{(3)} \]

### Step 4: Substitute Equation (3) into Equation (2)
Substituting \( a \) from equation (3) into equation (2):
\[ 6(1 - r)(1 + r) = \frac{9}{2} \]

### Step 5: Simplify the Equation
Expanding the left side:
\[ 6(1 - r^2) = \frac{9}{2} \]
\[ 36 - 36r^2 = 9 \quad \text{(Multiply both sides by 2 to eliminate the fraction)} \]
\[ 36r^2 = 36 - 9 \]
\[ 36r^2 = 27 \]
\[ r^2 = \frac{27}{36} = \frac{3}{4} \]
\[ r = \pm \frac{\sqrt{3}}{2} \]

### Step 6: Find \( a \) Using \( r \)
Using \( r = \frac{\sqrt{3}}{2} \) in equation (3):
\[ a = 6(1 - \frac{\sqrt{3}}{2}) = 6 \left(\frac{2 - \sqrt{3}}{2}\right) = 3(2 - \sqrt{3}) = 6 - 3\sqrt{3} \]

Using \( r = -\frac{\sqrt{3}}{2} \):
\[ a = 6(1 + \frac{\sqrt{3}}{2}) = 6 \left(\frac{2 + \sqrt{3}}{2}\right) = 3(2 + \sqrt{3}) = 6 + 3\sqrt{3} \]

### Step 7: Conclusion
Thus, the first term \( a \) can be either:
\[ a = 6 - 3\sqrt{3} \quad \text{or} \quad a = 6 + 3\sqrt{3} \]