In a G.P.sum of `2^(nd),3^(rd)` and `4^(th)` term is 3 and that `6^(th),7^(th)` and `8^(th)` term is 243 then `S_(50)=`

Options

A. `(3^(50)+1)/26`
B. `(3^(50)-1)/13`
C. `(3^(50)-1)/26`
D. `(3^(49)-1)/26`

Answer

To solve the problem, we need to find the sum of the first 50 terms of a geometric progression (G.P.) given certain conditions about its terms.

### Step-by-Step Solution:

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 terms of the G.P. can be expressed as:
- 2nd term: \( ar \)
- 3rd term: \( ar^2 \)
- 4th term: \( ar^3 \)

2. **Set Up the First Equation**:
According to the problem, the sum of the 2nd, 3rd, and 4th terms is given as 3:
\[ ar + ar^2 + ar^3 = 3 \]
Factoring out \( ar \):
\[ ar(1 + r + r^2) = 3 \quad \text{(Equation 1)} \]

3. **Define the 6th, 7th, and 8th Terms**:
The terms are:
- 6th term: \( ar^5 \)
- 7th term: \( ar^6 \)
- 8th term: \( ar^7 \)

4. **Set Up the Second Equation**:
The sum of the 6th, 7th, and 8th terms is given as 243:
\[ ar^5 + ar^6 + ar^7 = 243 \]
Factoring out \( ar^5 \):
\[ ar^5(1 + r + r^2) = 243 \quad \text{(Equation 2)} \]

5. **Divide Equation 2 by Equation 1**:
From Equation 2:
\[ ar^5(1 + r + r^2) = 243 \]
From Equation 1:
\[ ar(1 + r + r^2) = 3 \]
Dividing these two equations:
\[ \frac{ar^5(1 + r + r^2)}{ar(1 + r + r^2)} = \frac{243}{3} \]
This simplifies to:
\[ r^4 = 81 \]
Taking the fourth root:
\[ r = 3 \]

6. **Substitute \( r \) Back into Equation 1**:
Substitute \( r = 3 \) into Equation 1:
\[ ar(1 + 3 + 9) = 3 \]
Simplifying:
\[ ar \cdot 13 = 3 \implies ar = \frac{3}{13} \]

7. **Find \( a \)**:
Since \( r = 3 \):
\[ a \cdot 3 = \frac{3}{13} \implies a = \frac{1}{13} \]

8. **Find the Sum of the First 50 Terms**:
The formula for the sum of the first \( n \) terms of a G.P. is:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
For \( n = 50 \):
\[ S_{50} = \frac{1}{13} \cdot \frac{3^{50} - 1}{3 - 1} = \frac{1}{13} \cdot \frac{3^{50} - 1}{2} \]
Simplifying:
\[ S_{50} = \frac{3^{50} - 1}{26} \]

### Final Answer:
\[ S_{50} = \frac{3^{50} - 1}{26} \]