The first term of a G.P. in 27. If the `8^(th)` term be `(1)/(81)`, what will be the sum of 10 terms ?

Answer

To solve the problem, we need to find the sum of the first 10 terms of a geometric progression (G.P.) where the first term \( a = 27 \) and the 8th term \( a r^7 = \frac{1}{81} \).

### Step-by-step Solution:

1. **Identify the first term and the 8th term**:
- Given:
- First term \( a = 27 \)
- 8th term \( a r^7 = \frac{1}{81} \)

2. **Set up the equation for the 8th term**:
- From the formula for the \( n \)-th term of a G.P., we know:
\[ a r^{n-1} = \text{8th term} \]
- Therefore, we can write:
\[ 27 r^7 = \frac{1}{81} \]

3. **Solve for \( r^7 \)**:
- Rearranging the equation gives:
\[ r^7 = \frac{1}{81} \div 27 \]
- Since \( 27 = 3^3 \) and \( 81 = 3^4 \), we can write:
\[ r^7 = \frac{1}{3^4} \div 3^3 = \frac{1}{3^4} \times \frac{1}{3^3} = \frac{1}{3^{4+3}} = \frac{1}{3^7} \]
- Thus, we have:
\[ r^7 = \frac{1}{2187} \]

4. **Find \( r \)**:
- Taking the 7th root of both sides:
\[ r = \left(\frac{1}{3^7}\right)^{1/7} = \frac{1}{3} \]

5. **Find the sum of the first 10 terms**:
- The formula for the sum of the first \( n \) terms of a G.P. is:
\[ S_n = \frac{a(1 - r^n)}{1 - r} \]
- For \( n = 10 \):
\[ S_{10} = \frac{27(1 - (1/3)^{10})}{1 - (1/3)} \]
- Simplifying the denominator:
\[ 1 - (1/3) = \frac{2}{3} \]
- Thus:
\[ S_{10} = \frac{27(1 - (1/3)^{10})}{\frac{2}{3}} = 27 \cdot \frac{3}{2} (1 - (1/3)^{10}) \]
- This simplifies to:
\[ S_{10} = 40.5(1 - (1/3)^{10}) \]

6. **Calculate \( (1/3)^{10} \)**:
- \( (1/3)^{10} = \frac{1}{59049} \)
- Therefore:
\[ S_{10} = 40.5 \left(1 - \frac{1}{59049}\right) = 40.5 \left(\frac{59048}{59049}\right) \]

7. **Final Calculation**:
- The final sum \( S_{10} \) is approximately:
\[ S_{10} \approx 40.5 \cdot 0.999983 = 40.499 \]

### Final Answer:
The sum of the first 10 terms of the G.P. is approximately \( 40.5 \).