If the sum of the first two terms and the sum of the first four terms of a geometric progression with positive common ratio are 8 and 80 respectively, then what is the 6th term?

Options

A. 88
B. 243
C. 486
D. 1458

Explanation

To solve the problem, we need to find the 6th term of a geometric progression (GP) given the sums of the first two and four terms. Let's denote the first term of the GP as \( A \) and the common ratio as \( r \).

### Step-by-Step Solution:

1. **Write the equations for the sums:**
- The sum of the first two terms of the GP is given by:
\[ S_2 = A + Ar = A(1 + r) = 8 \]
- The sum of the first four terms of the GP is given by:
\[ S_4 = A + Ar + Ar^2 + Ar^3 = A(1 + r + r^2 + r^3) = 80 \]

2. **Express \( S_4 \) in terms of \( S_2 \):**
- We can factor \( S_4 \) using \( S_2 \):
\[ S_4 = A(1 + r)(1 + r^2) = 80 \]
- From the first equation, we know \( A(1 + r) = 8 \). Substitute this into the equation for \( S_4 \):
\[ 8(1 + r^2) = 80 \]
- Simplifying gives:
\[ 1 + r^2 = 10 \]
- Therefore:
\[ r^2 = 10 - 1 = 9 \]
- Taking the square root gives:
\[ r = 3 \quad (\text{since the common ratio is positive}) \]

3. **Find the value of \( A \):**
- Substitute \( r \) back into the first equation:
\[ A(1 + 3) = 8 \]
\[ A \cdot 4 = 8 \]
\[ A = \frac{8}{4} = 2 \]

4. **Calculate the 6th term of the GP:**
- The 6th term \( A_6 \) is given by:
\[ A_6 = A r^5 \]
- Substitute the values of \( A \) and \( r \):
\[ A_6 = 2 \cdot 3^5 \]
- Calculate \( 3^5 \):
\[ 3^5 = 243 \]
- Therefore:
\[ A_6 = 2 \cdot 243 = 486 \]

### Final Answer:
The 6th term of the geometric progression is \( \boxed{486} \).