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 G.P. is 3 and the sum of the squares of its terms is also 3.

### Step 1: Write the formula for 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:
\[ S = \frac{a}{1 - r} \]
According to the problem, the sum is 3:
\[ \frac{a}{1 - r} = 3 \]

### Step 2: Rearrange the equation to express \( a \)
From the equation above, we can express \( a \) in terms of \( r \):
\[ a = 3(1 - r) \]

### Step 3: Write the formula for the sum of the squares of the terms
The terms of the G.P. are \( a, ar, ar^2, ar^3, \ldots \). The squares of these terms are \( a^2, (ar)^2, (ar^2)^2, (ar^3)^2, \ldots \). The sum of the squares of the terms is:
\[ S_{\text{squares}} = \frac{a^2}{1 - r^2} \]
According to the problem, this sum is also 3:
\[ \frac{a^2}{1 - r^2} = 3 \]

### Step 4: Rearrange the equation to express \( a^2 \)
From the equation above, we can express \( a^2 \) in terms of \( r \):
\[ a^2 = 3(1 - r^2) \]

### Step 5: Substitute \( a \) from Step 2 into Step 4
Now we substitute \( a = 3(1 - r) \) into \( a^2 = 3(1 - r^2) \):
\[ (3(1 - r))^2 = 3(1 - r^2) \]
Expanding the left side:
\[ 9(1 - 2r + r^2) = 3(1 - r^2) \]
This simplifies to:
\[ 9 - 18r + 9r^2 = 3 - 3r^2 \]

### Step 6: Rearrange to form a quadratic equation
Rearranging gives:
\[ 9r^2 + 3r^2 - 18r + 9 - 3 = 0 \]
\[ 12r^2 - 18r + 6 = 0 \]

### Step 7: Simplify the quadratic equation
Dividing the entire equation by 6:
\[ 2r^2 - 3r + 1 = 0 \]

### Step 8: Solve the quadratic equation using the quadratic formula
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 2, b = -3, c = 1 \):
\[ r = \frac{3 \pm \sqrt{(-3)^2 - 4 \cdot 2 \cdot 1}}{2 \cdot 2} \]
\[ r = \frac{3 \pm \sqrt{9 - 8}}{4} \]
\[ r = \frac{3 \pm 1}{4} \]
This gives us two possible values for \( r \):
\[ r = \frac{4}{4} = 1 \quad \text{(not valid since } |r| < 1\text{)} \]
\[ r = \frac{2}{4} = \frac{1}{2} \]

### Step 9: Find the value of \( a \)
Now substituting \( r = \frac{1}{2} \) back into the equation for \( a \):
\[ a = 3(1 - \frac{1}{2}) = 3 \cdot \frac{1}{2} = \frac{3}{2} \]

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