The sum of first three terms of a G.P is 16 and the sum of the next three terms is 128. Determine the first term, the common ratio and the sum to n terms of the G.P.

Explanation

To solve the problem, we need to find the first term (a), the common ratio (r), and the sum to n terms of the given geometric progression (G.P.).

### Step-by-Step Solution:

1. **Understanding the Given Information:**
- The sum of the first three terms of the G.P. is given as \( S_3 = 16 \).
- The sum of the next three terms (i.e., the 4th, 5th, and 6th terms) is given as \( S_6 - S_3 = 128 \).

2. **Finding the Total Sum of the First Six Terms:**
- Since \( S_6 = S_3 + \text{(sum of next three terms)} \):
\[ S_6 = 16 + 128 = 144 \]

3. **Using the Formula for the Sum of the First n Terms of a G.P.:**
- The formula for the sum of the first n terms of a G.P. is:
\[ S_n = a \frac{1 - r^n}{1 - r} \]
- For \( n = 3 \):
\[ S_3 = a \frac{1 - r^3}{1 - r} = 16 \]
- For \( n = 6 \):
\[ S_6 = a \frac{1 - r^6}{1 - r} = 144 \]

4. **Setting Up the Equations:**
- From \( S_3 \):
\[ a(1 - r^3) = 16(1 - r) \quad \text{(1)} \]
- From \( S_6 \):
\[ a(1 - r^6) = 144(1 - r) \quad \text{(2)} \]

5. **Dividing Equation (2) by Equation (1):**
\[ \frac{a(1 - r^6)}{a(1 - r^3)} = \frac{144(1 - r)}{16(1 - r)} \]
- This simplifies to:
\[ \frac{1 - r^6}{1 - r^3} = \frac{144}{16} = 9 \]

6. **Simplifying the Left Side:**
- The expression \( 1 - r^6 \) can be factored as:
\[ 1 - r^6 = (1 - r^3)(1 + r^3) \]
- Thus, we have:
\[ 1 + r^3 = 9 \]
- Therefore:
\[ r^3 = 8 \quad \Rightarrow \quad r = 2 \]

7. **Substituting r Back to Find a:**
- Substitute \( r = 2 \) into Equation (1):
\[ a(1 - 2^3) = 16(1 - 2) \]
- This simplifies to:
\[ a(1 - 8) = 16(-1) \quad \Rightarrow \quad -7a = -16 \quad \Rightarrow \quad a = \frac{16}{7} \]

8. **Finding the Sum to n Terms:**
- Since \( r = 2 \) (which is greater than 1), the formula for the sum of the first n terms is:
\[ S_n = a \frac{r^n - 1}{r - 1} \]
- Substituting \( a \) and \( r \):
\[ S_n = \frac{16}{7} \frac{2^n - 1}{2 - 1} = \frac{16}{7} (2^n - 1) \]

### Final Answers:
- First term \( a = \frac{16}{7} \)
- Common ratio \( r = 2 \)
- Sum to n terms \( S_n = \frac{16}{7} (2^n - 1) \)