The `1^(st)` , `2^(nd)` and `3^(rd)` terms of an arithmetic series are `a`, `b` and `a^(2)` where `'a'` is negative. The `1^(st)`, `2^(nd)` and `3^(rd)` terms of a geometric series are `a`, `a^(2)` and `b` respectively.
The sum of infinite geometric series is

Options

A. `(-1)/(2)`
B. `(-3)/(2)`
C. `(-1)/(3)`
D. None of these

Explanation

To solve the problem, we need to find the sum of an infinite geometric series given the conditions of an arithmetic series and a geometric series. Let's break it down step by step.

### Step 1: Set up the equations from the arithmetic series
The first three terms of the arithmetic series are given as \( a \), \( b \), and \( a^2 \). For these terms to be in an arithmetic progression (AP), we have the condition:

\[ 2b = a + a^2 \]

### Step 2: Set up the equations from the geometric series
The first three terms of the geometric series are \( a \), \( a^2 \), and \( b \). For these terms to be in a geometric progression (GP), we have the condition:

\[ a^2 = \sqrt{a \cdot b} \]

Squaring both sides gives us:

\[ a^4 = ab \]

### Step 3: Solve the equations
Now we have two equations:

1. \( 2b = a + a^2 \) (Equation 1)
2. \( a^4 = ab \) (Equation 2)

From Equation 2, we can express \( b \) in terms of \( a \):

\[ b = \frac{a^4}{a} = a^3 \]

### Step 4: Substitute \( b \) in Equation 1
Substituting \( b = a^3 \) into Equation 1 gives:

\[ 2(a^3) = a + a^2 \]

This simplifies to:

\[ 2a^3 - a^2 - a = 0 \]

### Step 5: Factor the equation
Factoring out \( a \):

\[ a(2a^2 - a - 1) = 0 \]

This gives us \( a = 0 \) or \( 2a^2 - a - 1 = 0 \). Since \( a \) is negative, we will ignore \( a = 0 \) and solve the quadratic equation:

\[ 2a^2 - a - 1 = 0 \]

### Step 6: Use the quadratic formula
Using the quadratic formula \( a = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):

Here, \( a = 2 \), \( b = -1 \), and \( c = -1 \):

\[ a = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} \]
\[ = \frac{1 \pm \sqrt{1 + 8}}{4} \]
\[ = \frac{1 \pm 3}{4} \]

This gives us two potential solutions:

\[ a = 1 \quad \text{or} \quad a = -\frac{1}{2} \]

Since \( a \) is negative, we take \( a = -\frac{1}{2} \).

### Step 7: Find \( b \)
Now substituting \( a = -\frac{1}{2} \) back into the equation for \( b \):

\[ b = a^3 = \left(-\frac{1}{2}\right)^3 = -\frac{1}{8} \]

### Step 8: Write the geometric series
The first three terms of the geometric series are:

\[ -\frac{1}{2}, \left(-\frac{1}{2}\right)^2 = \frac{1}{4}, -\frac{1}{8} \]

### Step 9: Find the common ratio \( r \)
The common ratio \( r \) can be found as:

\[ r = \frac{\text{second term}}{\text{first term}} = \frac{\frac{1}{4}}{-\frac{1}{2}} = -\frac{1}{2} \]

### Step 10: Calculate the sum of the infinite geometric series
The sum \( S \) of an infinite geometric series is given by:

\[ S = \frac{a}{1 - r} \]

Here, \( a = -\frac{1}{2} \) and \( r = -\frac{1}{2} \):

\[ S = \frac{-\frac{1}{2}}{1 - (-\frac{1}{2})} = \frac{-\frac{1}{2}}{1 + \frac{1}{2}} = \frac{-\frac{1}{2}}{\frac{3}{2}} = -\frac{1}{3} \]

### Final Answer
The sum of the infinite geometric series is:

\[ \boxed{-\frac{1}{3}} \]