The fourth term, the seventh term and the last term of a geometric progression are 10, 80 and 2560 respectively. Find its first term, common ration and number of terms.

Answer

To solve the problem, we need to find the first term, common ratio, and number of terms of a geometric progression (GP) given that the fourth term is 10, the seventh term is 80, and the last term is 2560.

### Step-by-Step Solution:

1. **Understand the nth term formula for GP**:
The nth term of a geometric progression can be expressed as:
\[ T_n = a \cdot r^{n-1} \]
where \( a \) is the first term, \( r \) is the common ratio, and \( n \) is the term number.

2. **Set up equations for the given terms**:
- For the fourth term (\( T_4 \)):
\[ T_4 = a \cdot r^{4-1} = a \cdot r^3 = 10 \quad \text{(Equation 1)} \]
- For the seventh term (\( T_7 \)):
\[ T_7 = a \cdot r^{7-1} = a \cdot r^6 = 80 \quad \text{(Equation 2)} \]
- For the last term (\( T_n \)):
\[ T_n = a \cdot r^{n-1} = 2560 \quad \text{(Equation 3)} \]

3. **Divide Equation 2 by Equation 1** to eliminate \( a \):
\[ \frac{a \cdot r^6}{a \cdot r^3} = \frac{80}{10} \]
This simplifies to:
\[ r^3 = 8 \quad \Rightarrow \quad r = 2 \quad \text{(since } 2^3 = 8\text{)} \]

4. **Substitute \( r \) back into Equation 1** to find \( a \):
\[ a \cdot (2^3) = 10 \quad \Rightarrow \quad a \cdot 8 = 10 \quad \Rightarrow \quad a = \frac{10}{8} = \frac{5}{4} \]

5. **Now substitute \( a \) and \( r \) into Equation 3** to find \( n \):
\[ \frac{5}{4} \cdot (2^{n-1}) = 2560 \]
Multiply both sides by 4:
\[ 5 \cdot 2^{n-1} = 10240 \]
Divide both sides by 5:
\[ 2^{n-1} = 2048 \]
Since \( 2048 = 2^{11} \):
\[ n - 1 = 11 \quad \Rightarrow \quad n = 12 \]

### Final Results:
- First term \( a = \frac{5}{4} \)
- Common ratio \( r = 2 \)
- Number of terms \( n = 12 \)