If the sum of an infinite G.P. be 3 and the sum of the squares of its term is also 3, then its first term and common ratio are

Options

A. `3//2,1//2`
B. `1//2,3//2`
C. `1,1//2`
D. none of these

Answer

To solve the problem, we need to find the first term (a) and the common ratio (r) of an infinite geometric progression (G.P.) given that the sum of the series is 3 and the sum of the squares of its terms is also 3.

### Step-by-Step Solution:

1. **Understanding the Sum of an Infinite G.P.**:
The sum \( S \) of an infinite G.P. with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by the formula:
\[ S = \frac{a}{1 - r} \]
Given that \( S = 3 \), we can write:
\[ \frac{a}{1 - r} = 3 \quad \text{(Equation 1)} \]

2. **Sum of the Squares of the Terms**:
The terms of the G.P. are \( a, ar, ar^2, ar^3, \ldots \). The sum of the squares of these terms is:
\[ S_{\text{squares}} = a^2 + (ar)^2 + (ar^2)^2 + \ldots = a^2(1 + r^2 + r^4 + \ldots) \]
The sum of the series \( 1 + r^2 + r^4 + \ldots \) is also a G.P. with first term 1 and common ratio \( r^2 \):
\[ S_{\text{squares}} = a^2 \cdot \frac{1}{1 - r^2} \]
Given that this sum is also equal to 3, we can write:
\[ \frac{a^2}{1 - r^2} = 3 \quad \text{(Equation 2)} \]

3. **Solving the Equations**:
From Equation 1, we can express \( a \) in terms of \( r \):
\[ a = 3(1 - r) \]

4. **Substituting \( a \) into Equation 2**:
Substitute \( a \) into Equation 2:
\[ \frac{(3(1 - r))^2}{1 - r^2} = 3 \]
Simplifying this gives:
\[ \frac{9(1 - r)^2}{1 - r^2} = 3 \]
Cross-multiplying yields:
\[ 9(1 - r)^2 = 3(1 - r^2) \]
Expanding both sides:
\[ 9(1 - 2r + r^2) = 3(1 - r^2) \]
This simplifies to:
\[ 9 - 18r + 9r^2 = 3 - 3r^2 \]
Rearranging gives:
\[ 12r^2 - 18r + 6 = 0 \]
Dividing the entire equation by 6:
\[ 2r^2 - 3r + 1 = 0 \]

5. **Finding the Roots**:
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -3, c = 1 \):
\[ r = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \]
\[ r = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \]
Thus, the possible values for \( r \) are:
\[ r = 1 \quad \text{or} \quad r = \frac{1}{2} \]
Since \( |r| < 1 \), we take \( r = \frac{1}{2} \).

6. **Finding the First Term**:
Now substitute \( r = \frac{1}{2} \) back into Equation 1 to find \( a \):
\[ a = 3(1 - \frac{1}{2}) = 3 \cdot \frac{1}{2} = \frac{3}{2} \]

### Final Answer:
The first term \( a \) is \( \frac{3}{2} \) and the common ratio \( r \) is \( \frac{1}{2} \).