The fifth term of a G.P. is 81 whereas its second term is 24. Find the series and sum of its first eight terms.

Explanation

To solve the problem step by step, let's denote the first term of the geometric progression (G.P.) as \( a \) and the common ratio as \( r \).

### Step 1: Write down the formulas for the terms of the G.P.
The \( n \)-th term of a G.P. is given by:
\[ T_n = a \cdot r^{n-1} \]
From the problem, we know:
- The fifth term \( T_5 = 81 \)
- The second term \( T_2 = 24 \)

### Step 2: Set up equations based on the given terms
Using the formula for the terms:
\[ T_5 = a \cdot r^{4} = 81 \quad \text{(1)} \]
\[ T_2 = a \cdot r^{1} = 24 \quad \text{(2)} \]

### Step 3: Divide the equations to eliminate \( a \)
Dividing equation (1) by equation (2):
\[ \frac{T_5}{T_2} = \frac{a \cdot r^{4}}{a \cdot r^{1}} = \frac{81}{24} \]
This simplifies to:
\[ r^{3} = \frac{81}{24} \]

### Step 4: Simplify the fraction
Now simplify \( \frac{81}{24} \):
\[ \frac{81}{24} = \frac{27}{8} \quad \text{(by dividing both numerator and denominator by 3)} \]
Thus, we have:
\[ r^{3} = \frac{27}{8} \]

### Step 5: Find the value of \( r \)
Taking the cube root of both sides:
\[ r = \sqrt[3]{\frac{27}{8}} = \frac{3}{2} \]

### Step 6: Substitute \( r \) back to find \( a \)
Now substitute \( r \) back into equation (2) to find \( a \):
\[ a \cdot \frac{3}{2} = 24 \]
Solving for \( a \):
\[ a = 24 \cdot \frac{2}{3} = 16 \]

### Step 7: Write the series
Now we have \( a = 16 \) and \( r = \frac{3}{2} \). The first eight terms of the G.P. are:
\[ 16, \quad 16 \cdot \frac{3}{2} = 24, \quad 16 \cdot \left(\frac{3}{2}\right)^2 = 36, \quad 16 \cdot \left(\frac{3}{2}\right)^3 = 54, \quad 16 \cdot \left(\frac{3}{2}\right)^4 = 81, \quad 16 \cdot \left(\frac{3}{2}\right)^5 = 121.5, \quad 16 \cdot \left(\frac{3}{2}\right)^6 = 182.25, \quad 16 \cdot \left(\frac{3}{2}\right)^7 = 273.375 \]

### Step 8: Calculate the sum of the first eight terms
The sum of the first \( n \) terms of a G.P. is given by:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
For \( n = 8 \):
\[ S_8 = 16 \cdot \frac{\left(\frac{3}{2}\right)^8 - 1}{\frac{3}{2} - 1} \]
Calculating \( \left(\frac{3}{2}\right)^8 \):
\[ \left(\frac{3}{2}\right)^8 = \frac{6561}{256} \]
Thus,
\[ S_8 = 16 \cdot \frac{\frac{6561}{256} - 1}{\frac{1}{2}} = 16 \cdot \frac{\frac{6561 - 256}{256}}{\frac{1}{2}} = 16 \cdot \frac{6305}{256} \cdot 2 = \frac{20160}{256} = 78.75 \]

### Final Result
The series is:
\[ 16, 24, 36, 54, 81, 121.5, 182.25, 273.375 \]
The sum of the first eight terms is:
\[ S_8 = 78.75 \]