Find the G.P. whose 2nd and 5th terms are `-(3)/(2)" and "(81)/(16)` respectively.

Explanation

To find the geometric progression (G.P.) whose 2nd and 5th terms are \(-\frac{3}{2}\) and \(\frac{81}{16}\) respectively, we can follow these steps:

### Step 1: Use the formula for the nth term of a G.P.
The nth term of a geometric progression can be expressed as:
\[ a_n = a \cdot r^{n-1} \]
where \(a\) is the first term and \(r\) is the common ratio.

### Step 2: Write equations for the 2nd and 5th terms
From the problem, we know:
- The 2nd term \(a_2 = a \cdot r = -\frac{3}{2}\)
- The 5th term \(a_5 = a \cdot r^4 = \frac{81}{16}\)

### Step 3: Set up the equations
We have the following two equations:
1. \(a \cdot r = -\frac{3}{2}\) (Equation 1)
2. \(a \cdot r^4 = \frac{81}{16}\) (Equation 2)

### Step 4: Divide Equation 2 by Equation 1
To eliminate \(a\), we can divide Equation 2 by Equation 1:
\[ \frac{a \cdot r^4}{a \cdot r} = \frac{\frac{81}{16}}{-\frac{3}{2}} \]
This simplifies to:
\[ r^3 = \frac{81}{16} \cdot \left(-\frac{2}{3}\right) \]

### Step 5: Simplify the right side
Calculating the right side:
\[ r^3 = \frac{81 \cdot (-2)}{16 \cdot 3} = \frac{-162}{48} \]
Now simplify \(-162/48\):
\[ r^3 = -\frac{27}{8} \]

### Step 6: Solve for \(r\)
Taking the cube root of both sides:
\[ r = \sqrt[3]{-\frac{27}{8}} = -\frac{3}{2} \]

### Step 7: Substitute \(r\) back into Equation 1 to find \(a\)
Now substituting \(r\) back into Equation 1:
\[ a \cdot \left(-\frac{3}{2}\right) = -\frac{3}{2} \]
This gives:
\[ a = 1 \]

### Step 8: Write the terms of the G.P.
Now that we have \(a = 1\) and \(r = -\frac{3}{2}\), we can find the terms of the G.P.:
- First term \(a_1 = a = 1\)
- Second term \(a_2 = a \cdot r = 1 \cdot \left(-\frac{3}{2}\right) = -\frac{3}{2}\)
- Third term \(a_3 = a \cdot r^2 = 1 \cdot \left(-\frac{3}{2}\right)^2 = \frac{9}{4}\)
- Fourth term \(a_4 = a \cdot r^3 = 1 \cdot \left(-\frac{3}{2}\right)^3 = -\frac{27}{8}\)
- Fifth term \(a_5 = a \cdot r^4 = 1 \cdot \left(-\frac{3}{2}\right)^4 = \frac{81}{16}\)

### Final G.P. Terms
Thus, the G.P. is:
\[ 1, -\frac{3}{2}, \frac{9}{4}, -\frac{27}{8}, \frac{81}{16}, \ldots \]