The sum to infinity of a G.P. is 3 and the sum of squares of its terms is also 3. Find the G.P.

Explanation

To find the geometric progression (G.P.) given that the sum to infinity is 3 and the sum of squares of its terms is also 3, we can follow these steps:

### Step 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 terms of the G.P. can be expressed as:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)
- And so on.

### Step 2: Use the formula for the sum to infinity of a G.P.
The sum to infinity of a G.P. is given by the formula:
\[ S_{\infty} = \frac{a}{1 - r} \]
According to the problem, this sum is equal to 3:
\[ \frac{a}{1 - r} = 3 \]
From this, we can express \( a \) in terms of \( r \):
\[ a = 3(1 - r) \]

### Step 3: Use the formula for the sum of squares of the terms of the G.P.
The sum of squares of the terms of the G.P. is also a G.P. with the first term \( a^2 \) and common ratio \( r^2 \). The sum to infinity for this series is:
\[ S_{\infty} = \frac{a^2}{1 - r^2} \]
According to the problem, this sum is also equal to 3:
\[ \frac{a^2}{1 - r^2} = 3 \]
From this, we can express \( a^2 \) in terms of \( r \):
\[ a^2 = 3(1 - r^2) \]

### Step 4: Substitute \( a \) from Step 2 into Step 3
Now we substitute \( a = 3(1 - r) \) into the equation \( a^2 = 3(1 - r^2) \):
\[ (3(1 - r))^2 = 3(1 - r^2) \]
Expanding both sides gives:
\[ 9(1 - 2r + r^2) = 3(1 - r^2) \]
This simplifies to:
\[ 9 - 18r + 9r^2 = 3 - 3r^2 \]
Rearranging terms leads to:
\[ 12r^2 - 18r + 6 = 0 \]

### Step 5: Simplify the quadratic equation
Dividing the entire equation by 6 gives:
\[ 2r^2 - 3r + 1 = 0 \]

### Step 6: Solve the quadratic equation using the quadratic formula
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2 \), \( b = -3 \), and \( c = 1 \):
\[ r = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \]
Calculating the discriminant:
\[ r = \frac{3 \pm \sqrt{9 - 8}}{4} = \frac{3 \pm 1}{4} \]
This gives us two possible values for \( r \):
1. \( r = \frac{4}{4} = 1 \)
2. \( r = \frac{2}{4} = \frac{1}{2} \)

### Step 7: Find the corresponding values of \( a \)
Since \( r = 1 \) is not valid for a converging G.P., we take \( r = \frac{1}{2} \):
Substituting \( r \) back into the equation for \( a \):
\[ a = 3(1 - \frac{1}{2}) = 3 \cdot \frac{1}{2} = \frac{3}{2} \]

### Step 8: Write the terms of the G.P.
Now we can write the terms of the G.P.:
- First term: \( a = \frac{3}{2} \)
- Second term: \( ar = \frac{3}{2} \cdot \frac{1}{2} = \frac{3}{4} \)
- Third term: \( ar^2 = \frac{3}{2} \cdot \left(\frac{1}{2}\right)^2 = \frac{3}{8} \)

Thus, the G.P. is:
\[ \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \ldots \]

### Final Answer
The geometric progression is:
\[ \frac{3}{2}, \frac{3}{4}, \frac{3}{8}, \ldots \]