The sum of the terms of an infinite geometric progression is 3 and the sum of the squares of the terms is 81. Find the first term of the series.

Options

A. 5
B. `(27)/(5)`
C. `(31)/(6)`
D. `(19)/(3)`

Explanation

To solve the problem step by step, we will use the properties of geometric progressions (GP).

### Step 1: Understand the formulas for the sums
The sum \( S \) of an infinite geometric progression can be expressed as:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. We are given that \( S = 3 \).

### Step 2: Set up the first equation
From the information provided, we can set up our first equation:
\[ \frac{a}{1 - r} = 3 \quad \text{(Equation 1)} \]

### Step 3: Set up the second equation for the sum of squares
The sum of the squares of the terms in the GP is given by:
\[ S_{squares} = \frac{a^2}{1 - r^2} \]
We are given that this sum equals 81, so we can set up our second equation:
\[ \frac{a^2}{1 - r^2} = 81 \quad \text{(Equation 2)} \]

### Step 4: Solve Equation 1 for \( a \)
From Equation 1, we can express \( a \) in terms of \( r \):
\[ a = 3(1 - r) \quad \text{(Equation 3)} \]

### Step 5: Substitute Equation 3 into Equation 2
Now, substitute Equation 3 into Equation 2:
\[ \frac{(3(1 - r))^2}{1 - r^2} = 81 \]
This simplifies to:
\[ \frac{9(1 - r)^2}{1 - r^2} = 81 \]

### Step 6: Simplify the equation
Cross-multiply to eliminate the fraction:
\[ 9(1 - r)^2 = 81(1 - r^2) \]
Expanding both sides:
\[ 9(1 - 2r + r^2) = 81(1 - r^2) \]
This simplifies to:
\[ 9 - 18r + 9r^2 = 81 - 81r^2 \]
Rearranging gives:
\[ 90r^2 - 18r - 72 = 0 \]

### Step 7: Divide the entire equation by 18
To simplify, divide the entire equation by 18:
\[ 5r^2 - r - 4 = 0 \]

### Step 8: Factor the quadratic equation
Now, we can factor the quadratic:
\[ (5r + 8)(r - 1) = 0 \]
Setting each factor to zero gives:
\[ r = -\frac{8}{5} \quad \text{or} \quad r = 1 \]
Since \( r \) must be less than 1 for the GP to converge, we take \( r = -\frac{8}{5} \).

### Step 9: Substitute \( r \) back to find \( a \)
Now substitute \( r \) back into Equation 3:
\[ a = 3(1 - (-\frac{8}{5})) = 3(1 + \frac{8}{5}) = 3(\frac{5 + 8}{5}) = 3(\frac{13}{5}) = \frac{39}{5} \]

### Final Answer
Thus, the first term \( a \) of the series is:
\[ \boxed{\frac{39}{5}} \]