The sum of first three terms of a G.P. is `(39)/(10)` and their product is 1. Find the common ratio and the terms.

Explanation

To solve the problem, we need to find the first three terms of a geometric progression (G.P.) given that their sum is \( \frac{39}{10} \) and their product is \( 1 \).

### Step-by-step solution:

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:
\[ \frac{a}{r}, \quad a, \quad ar \]

2. **Write the equation for the product**:
According to the problem, the product of these three terms is \( 1 \):
\[ \left(\frac{a}{r}\right) \cdot a \cdot (ar) = 1 \]
Simplifying this gives:
\[ \frac{a^3}{r} = 1 \implies a^3 = r \]

3. **Write the equation for the sum**:
The sum of the three terms is given as \( \frac{39}{10} \):
\[ \frac{a}{r} + a + ar = \frac{39}{10} \]
Substituting \( a \) from the product equation:
\[ \frac{a}{r} + a + ar = \frac{39}{10} \]
We can express \( a \) in terms of \( r \):
\[ a = r^{1/3} \]
Thus, substituting \( a \) into the sum equation:
\[ \frac{r^{1/3}}{r} + r^{1/3} + r^{1/3} \cdot r = \frac{39}{10} \]
Simplifying this gives:
\[ \frac{1}{r^{2/3}} + r^{1/3} + r^{4/3} = \frac{39}{10} \]

4. **Multiply through by \( 10r^{2/3} \)** to eliminate the fraction:
\[ 10 + 10r^{5/3} + 10r^{2/3} = 39r^{2/3} \]
Rearranging gives:
\[ 10r^{5/3} - 29r^{2/3} + 10 = 0 \]

5. **Let \( x = r^{1/3} \)**, then \( r = x^3 \):
Substitute \( r \) in terms of \( x \):
\[ 10x^5 - 29x^2 + 10 = 0 \]

6. **Factor the polynomial**:
We can factor this polynomial:
\[ (2x^2 - 5)(5x^3 - 2) = 0 \]
Setting each factor to zero gives:
\[ 2x^2 - 5 = 0 \implies x^2 = \frac{5}{2} \implies x = \sqrt{\frac{5}{2}} \quad \text{(not valid since \(x\) must be real)} \]
\[ 5x^3 - 2 = 0 \implies x^3 = \frac{2}{5} \implies x = \sqrt[3]{\frac{2}{5}} \]

7. **Finding \( r \)**:
Now substituting back to find \( r \):
\[ r = x^3 = \frac{2}{5} \]

8. **Finding \( a \)**:
Using \( a^3 = r \):
\[ a^3 = \frac{2}{5} \implies a = \sqrt[3]{\frac{2}{5}} \]

9. **Finding the terms**:
The three terms of the G.P. are:
\[ \frac{a}{r}, \quad a, \quad ar \]
Substituting \( a \) and \( r \):
\[ \frac{\sqrt[3]{\frac{2}{5}}}{\frac{2}{5}}, \quad \sqrt[3]{\frac{2}{5}}, \quad \sqrt[3]{\frac{2}{5}} \cdot \frac{2}{5} \]
Simplifying gives:
\[ \frac{5\sqrt[3]{\frac{2}{5}}}{2}, \quad \sqrt[3]{\frac{2}{5}}, \quad \frac{2\sqrt[3]{\frac{2}{5}}}{5} \]

### Final Answer:
The common ratio \( r \) is \( \frac{2}{5} \) and the terms of the G.P. are \( \frac{5\sqrt[3]{\frac{2}{5}}}{2}, \sqrt[3]{\frac{2}{5}}, \frac{2\sqrt[3]{\frac{2}{5}}}{5} \).