The 4th term of a G.P. is 16 and 7th term is 128. Find the first term and common ration of the series.

Answer

To solve the problem of finding the first term and common ratio of a geometric progression (G.P.) where the 4th term is 16 and the 7th term is 128, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the General Formula for G.P.:**
The n-th term of a geometric progression can be expressed as:
\[ a_n = a \cdot r^{n-1} \]
where \( a \) is the first term and \( r \) is the common ratio.

2. **Set Up Equations Based on Given Terms:**
From the problem, we know:
- The 4th term \( a_4 = 16 \):
\[ a \cdot r^{4-1} = a \cdot r^3 = 16 \quad \text{(Equation 1)} \]
- The 7th term \( a_7 = 128 \):
\[ a \cdot r^{7-1} = a \cdot r^6 = 128 \quad \text{(Equation 2)} \]

3. **Divide Equation 2 by Equation 1:**
To eliminate \( a \), divide Equation 2 by Equation 1:
\[ \frac{a \cdot r^6}{a \cdot r^3} = \frac{128}{16} \]
This simplifies to:
\[ r^{6-3} = \frac{128}{16} \]
\[ r^3 = 8 \]

4. **Solve for the Common Ratio \( r \):**
Taking the cube root of both sides:
\[ r = 2 \]

5. **Substitute \( r \) back to find \( a \):**
Now, substitute \( r \) back into Equation 1 to find \( a \):
\[ a \cdot r^3 = 16 \]
\[ a \cdot 2^3 = 16 \]
\[ a \cdot 8 = 16 \]
\[ a = \frac{16}{8} = 2 \]

### Final Results:
- The first term \( a \) is 2.
- The common ratio \( r \) is 2.