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 solve the problem, we need to find the first term \( a \) 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.

### Step-by-Step Solution:

1. **Understanding the 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:
\[ S = \frac{a}{1 - r} \]
According to the problem, this sum is equal to 5:
\[ \frac{a}{1 - r} = 5 \quad \text{(1)} \]

2. **Sum of the Squares of the Terms**:
The sum of the squares of the terms of the G.P. is given by:
\[ S_{squares} = \frac{a^2}{1 - r^2} \]
This sum is also equal to 5:
\[ \frac{a^2}{1 - r^2} = 5 \quad \text{(2)} \]

3. **Expressing \( a \) from Equation (1)**:
From equation (1), we can express \( a \) in terms of \( r \):
\[ a = 5(1 - r) \quad \text{(3)} \]

4. **Substituting \( a \) into Equation (2)**:
Substitute equation (3) into equation (2):
\[ \frac{(5(1 - r))^2}{1 - r^2} = 5 \]
Simplifying this gives:
\[ \frac{25(1 - r)^2}{1 - r^2} = 5 \]

5. **Cross-Multiplying**:
Cross-multiply to eliminate the fraction:
\[ 25(1 - r)^2 = 5(1 - r^2) \]

6. **Expanding Both Sides**:
Expanding both sides:
\[ 25(1 - 2r + r^2) = 5(1 - r^2) \]
This simplifies to:
\[ 25 - 50r + 25r^2 = 5 - 5r^2 \]

7. **Rearranging the Equation**:
Rearranging gives:
\[ 30r^2 - 50r + 20 = 0 \]

8. **Dividing by 10**:
Simplifying the equation by dividing by 10:
\[ 3r^2 - 5r + 2 = 0 \]

9. **Factoring the Quadratic**:
Factoring the quadratic equation:
\[ (3r - 2)(r - 1) = 0 \]
Thus, the solutions for \( r \) are:
\[ r = \frac{2}{3} \quad \text{or} \quad r = 1 \]
Since \( |r| < 1 \), we take \( r = \frac{2}{3} \).

10. **Finding \( a \)**:
Substitute \( r \) back into equation (3) to find \( a \):
\[ a = 5(1 - \frac{2}{3}) = 5 \times \frac{1}{3} = \frac{5}{3} \]

### Final Answer:
The first term \( a \) of the infinite geometric progression is:
\[ \boxed{\frac{5}{3}} \]