The sum of the first three terms of a G.P. is `(13)/(12)` and their product is -1. Find the G.P.

Answer

To solve the problem, we need to find the first three terms of a geometric progression (G.P.) given that their sum is \( \frac{13}{12} \) and their product is \(-1\).

### Step 1: Define the terms of the G.P.
Let the first term be \( a \) and the common ratio be \( r \). The first three terms of the G.P. can be expressed as:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)

### Step 2: Set up the equations based on the given information
From the problem, we have two conditions:
1. The sum of the terms:
\[ a + ar + ar^2 = \frac{13}{12} \]
This can be factored as:
\[ a(1 + r + r^2) = \frac{13}{12} \]

2. The product of the terms:
\[ a \cdot ar \cdot ar^2 = -1 \]
This simplifies to:
\[ a^3 r^3 = -1 \]
Taking the cube root gives:
\[ ar = -1 \]

### Step 3: Substitute \( ar \) into the sum equation
From \( ar = -1 \), we can express \( a \) in terms of \( r \):
\[ a = \frac{-1}{r} \]

Substituting \( a \) into the sum equation:
\[ \frac{-1}{r}(1 + r + r^2) = \frac{13}{12} \]
Multiplying both sides by \(-r\) gives:
\[ 1 + r + r^2 = -\frac{13r}{12} \]
Rearranging this leads to:
\[ 12 + 12r + 12r^2 + 13r = 0 \]
This simplifies to:
\[ 12r^2 + 25r + 12 = 0 \]

### Step 4: Solve the quadratic equation
We can use the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
Here, \( a = 12 \), \( b = 25 \), and \( c = 12 \).
Calculating the discriminant:
\[ D = b^2 - 4ac = 25^2 - 4 \cdot 12 \cdot 12 = 625 - 576 = 49 \]
Now applying the quadratic formula:
\[ r = \frac{-25 \pm \sqrt{49}}{2 \cdot 12} = \frac{-25 \pm 7}{24} \]
Calculating the two possible values for \( r \):
1. \( r = \frac{-18}{24} = -\frac{3}{4} \)
2. \( r = \frac{-32}{24} = -\frac{4}{3} \)

### Step 5: Find the corresponding values of \( a \)
For \( r = -\frac{3}{4} \):
\[ ar = -1 \Rightarrow a \left(-\frac{3}{4}\right) = -1 \Rightarrow a = \frac{4}{3} \]
The terms are:
- First term: \( \frac{4}{3} \)
- Second term: \( -1 \)
- Third term: \( \frac{4}{3} \cdot -\frac{3}{4} = \frac{3}{4} \)

For \( r = -\frac{4}{3} \):
\[ ar = -1 \Rightarrow a \left(-\frac{4}{3}\right) = -1 \Rightarrow a = \frac{3}{4} \]
The terms are:
- First term: \( \frac{3}{4} \)
- Second term: \( -1 \)
- Third term: \( \frac{3}{4} \cdot -\frac{4}{3} = -4 \)

### Conclusion
Thus, the two possible G.P.s are:
1. \( \frac{4}{3}, -1, \frac{3}{4} \)
2. \( \frac{3}{4}, -1, \frac{4}{3} \)