Fourth and seventh terms of a G.P. are `(1)/(18)` and `-(1)/(486)` respectively. Find the G.P.

Answer

To find the geometric progression (G.P.) given the fourth and seventh terms, we can follow these steps:

### Step 1: Define the terms of the G.P.
Let the first term of the G.P. be \( a \) and the common ratio be \( r \). The \( n \)-th term of a G.P. is given by:
\[ T_n = a \cdot r^{n-1} \]
From the problem, we know:
- The fourth term \( T_4 = \frac{1}{18} \)
- The seventh term \( T_7 = -\frac{1}{486} \)

### Step 2: Write equations for the fourth and seventh terms.
Using the formula for the \( n \)-th term:
\[ T_4 = a \cdot r^{3} = \frac{1}{18} \quad \text{(Equation 1)} \]
\[ T_7 = a \cdot r^{6} = -\frac{1}{486} \quad \text{(Equation 2)} \]

### Step 3: Divide Equation 2 by Equation 1.
To eliminate \( a \), we can divide Equation 2 by Equation 1:
\[ \frac{T_7}{T_4} = \frac{a \cdot r^{6}}{a \cdot r^{3}} = \frac{-\frac{1}{486}}{\frac{1}{18}} \]
This simplifies to:
\[ r^{3} = \frac{-\frac{1}{486}}{\frac{1}{18}} = -\frac{1}{486} \cdot \frac{18}{1} = -\frac{18}{486} = -\frac{1}{27} \]

### Step 4: Solve for \( r \).
Now, we can solve for \( r \):
\[ r^{3} = -\frac{1}{27} \]
Taking the cube root of both sides:
\[ r = -\frac{1}{3} \]

### Step 5: Substitute \( r \) back into Equation 1 to find \( a \).
Now, substitute \( r \) back into Equation 1 to find \( a \):
\[ \frac{1}{18} = a \cdot \left(-\frac{1}{3}\right)^{3} \]
Calculating \( \left(-\frac{1}{3}\right)^{3} = -\frac{1}{27} \):
\[ \frac{1}{18} = a \cdot \left(-\frac{1}{27}\right) \]
Now, solve for \( a \):
\[ a = \frac{1}{18} \cdot \left(-27\right) = -\frac{27}{18} = -\frac{3}{2} \]

### Step 6: Write the G.P.
Now that we have \( a \) and \( r \), we can write the G.P.:
The terms of the G.P. are:
- First term: \( a = -\frac{3}{2} \)
- Second term: \( a \cdot r = -\frac{3}{2} \cdot -\frac{1}{3} = \frac{1}{2} \)
- Third term: \( a \cdot r^{2} = -\frac{3}{2} \cdot \left(-\frac{1}{3}\right)^{2} = -\frac{3}{2} \cdot \frac{1}{9} = -\frac{1}{6} \)
- Fourth term: \( T_4 = \frac{1}{18} \)
- Fifth term: \( T_5 = -\frac{1}{54} \)
- Sixth term: \( T_6 = \frac{1}{162} \)
- Seventh term: \( T_7 = -\frac{1}{486} \)

Thus, the G.P. is:
\[ -\frac{3}{2}, \frac{1}{2}, -\frac{1}{6}, \frac{1}{18}, -\frac{1}{54}, \frac{1}{162}, -\frac{1}{486} \]