The sum to infinity of a G.P. is 15 and the sum of squares of its terms is 45. Find the G.P

Explanation

To solve the problem, we need to find the terms of a geometric progression (G.P.) given the sum to infinity and the sum of squares of its terms.

### Step-by-Step Solution:

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 represented as \( a, ar, ar^2, ar^3, \ldots \).

2. **Sum to Infinity of the G.P.**:
The formula for the sum to infinity of a G.P. is given by:
\[ S = \frac{a}{1 - r} \]
We are given that this sum is equal to 15:
\[ \frac{a}{1 - r} = 15 \quad \text{(Equation 1)} \]

3. **Sum of Squares of the Terms**:
The terms of the G.P. squared are \( a^2, (ar)^2, (ar^2)^2, \ldots \). The sum of squares can be expressed as:
\[ S_{squares} = \frac{a^2}{1 - r^2} \]
We are given that this sum is equal to 45:
\[ \frac{a^2}{1 - r^2} = 45 \quad \text{(Equation 2)} \]

4. **Express \( 1 - r^2 \)**:
We can express \( 1 - r^2 \) in terms of \( 1 - r \):
\[ 1 - r^2 = (1 - r)(1 + r) \]
Substituting this into Equation 2 gives:
\[ \frac{a^2}{(1 - r)(1 + r)} = 45 \]

5. **Substituting Equation 1 into Equation 2**:
From Equation 1, we can express \( a \):
\[ a = 15(1 - r) \]
Substitute \( a \) into the modified Equation 2:
\[ \frac{(15(1 - r))^2}{(1 - r)(1 + r)} = 45 \]
Simplifying this gives:
\[ \frac{225(1 - r)}{1 + r} = 45 \]

6. **Cross-Multiplying**:
Cross-multiplying gives:
\[ 225(1 - r) = 45(1 + r) \]
Expanding both sides:
\[ 225 - 225r = 45 + 45r \]

7. **Combining Like Terms**:
Rearranging the equation gives:
\[ 225 - 45 = 225r + 45r \]
This simplifies to:
\[ 180 = 270r \]
Thus, solving for \( r \):
\[ r = \frac{180}{270} = \frac{2}{3} \]

8. **Finding \( a \)**:
Now substituting \( r \) back into Equation 1 to find \( a \):
\[ a = 15(1 - \frac{2}{3}) = 15 \cdot \frac{1}{3} = 5 \]

9. **Final G.P. Terms**:
The terms of the G.P. are:
- First term: \( a = 5 \)
- Second term: \( ar = 5 \cdot \frac{2}{3} = \frac{10}{3} \)
- Third term: \( ar^2 = 5 \cdot \left(\frac{2}{3}\right)^2 = 5 \cdot \frac{4}{9} = \frac{20}{9} \)
- And so on...

### Conclusion:
The G.P. is \( 5, \frac{10}{3}, \frac{20}{9}, \ldots \)