To solve the problem, we need to find the terms of a geometric progression (G.P.) given that the sum of the infinite series is 15 and the sum of the squares of these terms is 45.
### Step-by-Step Solution:
1. **Understanding the Infinite G.P.**:
The sum \( S \) of an infinite geometric series with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by:
\[
S = \frac{a}{1 - r}
\]
From the problem, we know:
\[
S = 15 \implies \frac{a}{1 - r} = 15 \quad \text{(Equation 1)}
\]
2. **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 \( a^2, a^2r^2, a^2r^4, a^2r^6, \ldots \). This forms another infinite G.P. with first term \( a^2 \) and common ratio \( r^2 \):
\[
S_{squares} = \frac{a^2}{1 - r^2}
\]
From the problem, we know:
\[
S_{squares} = 45 \implies \frac{a^2}{1 - r^2} = 45 \quad \text{(Equation 2)}
\]
3. **Expressing \( 1 - r^2 \)**:
We can rewrite \( 1 - r^2 \) using the identity \( 1 - r^2 = (1 - r)(1 + r) \). Thus, Equation 2 becomes:
\[
\frac{a^2}{(1 - r)(1 + r)} = 45
\]
4. **Substituting Equation 1 into Equation 2**:
From Equation 1, we have \( 1 - r = \frac{a}{15} \). Substituting this into Equation 2 gives:
\[
\frac{a^2}{\frac{a}{15}(1 + r)} = 45
\]
Simplifying this:
\[
\frac{15a}{1 + r} = 45 \implies 15a = 45(1 + r) \implies a = 3(1 + r) \quad \text{(Equation 3)}
\]
5. **Substituting \( a \) back**:
Substitute \( a \) from Equation 3 back into Equation 1:
\[
\frac{3(1 + r)}{1 - r} = 15
\]
Cross-multiplying gives:
\[
3(1 + r) = 15(1 - r) \implies 3 + 3r = 15 - 15r
\]
Rearranging terms:
\[
3r + 15r = 15 - 3 \implies 18r = 12 \implies r = \frac{2}{3}
\]
6. **Finding \( a \)**:
Now substitute \( r \) back into Equation 3 to find \( a \):
\[
a = 3(1 + \frac{2}{3}) = 3 \times \frac{5}{3} = 5
\]
7. **Finding the G.P. Terms**:
The first term \( a = 5 \) and the common ratio \( r = \frac{2}{3} \). The terms of the G.P. are:
- First term: \( a = 5 \)
- Second term: \( ar = 5 \times \frac{2}{3} = \frac{10}{3} \)
- Third term: \( ar^2 = 5 \times \left(\frac{2}{3}\right)^2 = 5 \times \frac{4}{9} = \frac{20}{9} \)
- Fourth term: \( ar^3 = 5 \times \left(\frac{2}{3}\right)^3 = 5 \times \frac{8}{27} = \frac{40}{27} \)
Thus, the G.P. is:
\[
5, \frac{10}{3}, \frac{20}{9}, \frac{40}{27}, \ldots
\]
### Final Answer:
The G.P. is \( 5, \frac{10}{3}, \frac{20}{9}, \frac{40}{27}, \ldots \)
