The 4th, 7th and last terms of a G.P. are 10,80 and 2560 respectively. Find the number of terms of the G.P.

Explanation

To solve the problem, we need to find the number of terms in a geometric progression (G.P.) where the 4th term is 10, the 7th term is 80, and the last term is 2560.

### Step-by-Step Solution:

1. **Define the terms of the G.P.**:
- Let the first term be \( a \) and the common ratio be \( r \).
- The \( n \)-th term of a G.P. can be expressed as:
\[ a_n = a \cdot r^{n-1} \]
- Therefore, we can write the given terms as:
- 4th term: \( a \cdot r^3 = 10 \) (Equation 1)
- 7th term: \( a \cdot r^6 = 80 \) (Equation 2)
- Last term: \( a \cdot r^{n-1} = 2560 \) (Equation 3)

2. **Divide Equation 2 by Equation 1**:
\[ \frac{a \cdot r^6}{a \cdot r^3} = \frac{80}{10} \]
This simplifies to:
\[ r^3 = 8 \]
Therefore, we find:
\[ r = 2 \]

3. **Substitute \( r \) back into Equation 1**:
\[ a \cdot (2^3) = 10 \]
Simplifying gives:
\[ a \cdot 8 = 10 \implies a = \frac{10}{8} = \frac{5}{4} \]

4. **Use \( a \) and \( r \) in Equation 3**:
Substitute \( a \) and \( r \) into Equation 3:
\[ \frac{5}{4} \cdot 2^{n-1} = 2560 \]
Multiply both sides by 4:
\[ 5 \cdot 2^{n-1} = 10240 \]
Now divide both sides by 5:
\[ 2^{n-1} = \frac{10240}{5} = 2048 \]

5. **Express 2048 as a power of 2**:
We know that:
\[ 2048 = 2^{11} \]
Therefore, we have:
\[ n - 1 = 11 \implies n = 12 \]

### Conclusion:
The number of terms in the G.P. is \( n = 12 \).