The sum of an infinite series is 15 and the sum of the squares of these terms is 45. Find the series.

Answer

To find the infinite series given that the sum is 15 and the sum of the squares of the terms is 45, we can follow these steps:

### Step 1: Define the Series
Assume the series is a geometric progression (GP) with first term \( a \) and common ratio \( r \). The terms of the series can be expressed as:
\[ a, ar, ar^2, ar^3, \ldots \]

### Step 2: Use the Sum of the Infinite Series
The formula for the sum of an infinite geometric series is given by:
\[ S = \frac{a}{1 - r} \]
According to the problem, the sum of the series is 15:
\[ \frac{a}{1 - r} = 15 \]
This is our **Equation 1**.

### Step 3: Use the Sum of the Squares of the Terms
The terms of the squares of the series are:
\[ a^2, a^2r^2, a^2r^4, a^2r^6, \ldots \]
This is also a geometric series with first term \( a^2 \) and common ratio \( r^2 \). The sum of the squares is given by:
\[ S_{squares} = \frac{a^2}{1 - r^2} \]
According to the problem, this sum is 45:
\[ \frac{a^2}{1 - r^2} = 45 \]
This is our **Equation 2**.

### Step 4: Solve Equation 1 for \( a \)
From Equation 1:
\[ a = 15(1 - r) \]

### Step 5: Substitute \( a \) in Equation 2
Substituting \( a \) in Equation 2:
\[ \frac{(15(1 - r))^2}{1 - r^2} = 45 \]
Expanding this gives:
\[ \frac{225(1 - 2r + r^2)}{1 - r^2} = 45 \]

### Step 6: Simplify the Equation
Cross-multiplying gives:
\[ 225(1 - 2r + r^2) = 45(1 - r^2) \]
Expanding both sides:
\[ 225 - 450r + 225r^2 = 45 - 45r^2 \]
Combining like terms:
\[ 225 + 45 = 450r + 225r^2 + 45r^2 \]
This simplifies to:
\[ 270r^2 - 450r + 180 = 0 \]

### Step 7: Solve the Quadratic Equation
Dividing the entire equation by 90:
\[ 3r^2 - 5r + 2 = 0 \]
Now we can use the quadratic formula:
\[ r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \( a = 3, b = -5, c = 2 \):
\[ r = \frac{5 \pm \sqrt{(-5)^2 - 4 \cdot 3 \cdot 2}}{2 \cdot 3} \]
\[ r = \frac{5 \pm \sqrt{25 - 24}}{6} \]
\[ r = \frac{5 \pm 1}{6} \]
Thus, we have two possible values for \( r \):
1. \( r = 1 \)
2. \( r = \frac{2}{3} \)

Since \( |r| < 1 \), we take \( r = \frac{2}{3} \).

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

### Step 9: Write the Series
The series is:
\[ 5, \frac{10}{3}, \frac{20}{9}, \frac{40}{27}, \ldots \]

### Final Answer
The series is:
\[ 5, \frac{10}{3}, \frac{20}{9}, \frac{40}{27}, \ldots \]