If the sum of an infinite G.P and the sum of the squares of its terms are both equal to 5 , then the first term is

Answer

To find the first term of an infinite geometric progression (G.P.) given that both the sum of the G.P. and the sum of the squares of its terms are equal to 5, 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. will be \( a, ar, ar^2, ar^3, \ldots \).

### Step 2: Write the formula for the sum of an infinite G.P.
The sum \( S_1 \) of an infinite G.P. is given by the formula:
\[ S_1 = \frac{a}{1 - r} \]
Given that \( S_1 = 5 \), we have:
\[ \frac{a}{1 - r} = 5 \quad \text{(1)} \]

### Step 3: Write the formula for the sum of the squares of the terms of the G.P.
The terms of the squares of the G.P. will be \( a^2, a^2r^2, a^2r^4, \ldots \). The sum \( S_2 \) of the squares of the terms is given by:
\[ S_2 = \frac{a^2}{1 - r^2} \]
Given that \( S_2 = 5 \), we have:
\[ \frac{a^2}{1 - r^2} = 5 \quad \text{(2)} \]

### Step 4: Solve the equations
From equation (1):
\[ a = 5(1 - r) \quad \text{(3)} \]

Substituting equation (3) into equation (2):
\[ \frac{(5(1 - r))^2}{1 - r^2} = 5 \]
This simplifies to:
\[ \frac{25(1 - r)^2}{1 - r^2} = 5 \]

### Step 5: Cross-multiply and simplify
Cross-multiplying gives:
\[ 25(1 - r)^2 = 5(1 - r^2) \]
Dividing both sides by 5:
\[ 5(1 - r)^2 = 1 - r^2 \]
Expanding both sides:
\[ 5(1 - 2r + r^2) = 1 - r^2 \]
This simplifies to:
\[ 5 - 10r + 5r^2 = 1 - r^2 \]
Rearranging gives:
\[ 6r^2 - 10r + 4 = 0 \]

### Step 6: Solve the quadratic equation
Using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ r = \frac{10 \pm \sqrt{(-10)^2 - 4 \cdot 6 \cdot 4}}{2 \cdot 6} \]
Calculating the discriminant:
\[ r = \frac{10 \pm \sqrt{100 - 96}}{12} = \frac{10 \pm 2}{12} \]
This gives two possible values for \( r \):
\[ r = \frac{12}{12} = 1 \quad \text{(not valid for G.P.)} \]
\[ r = \frac{8}{12} = \frac{2}{3} \]

### Step 7: Find the first term \( a \)
Substituting \( r = \frac{2}{3} \) back into equation (3):
\[ a = 5(1 - \frac{2}{3}) = 5 \cdot \frac{1}{3} = \frac{5}{3} \]

### Conclusion
The first term of the G.P. is:
\[ \boxed{\frac{5}{3}} \]