The `4^(th), 6^(th)` and the last term of a geometric progression are 10, 40 and 640 respectively. If the common ratio is positive, find the first term, common ratio and the number of terms of the series.

Answer

To solve the problem, we need to find the first term, common ratio, and the number of terms in a geometric progression (GP) where the 4th term is 10, the 6th term is 40, and the last term is 640.

### Step-by-Step Solution:

1. **Define the Variables:**
Let \( a \) be the first term and \( r \) be the common ratio of the GP.

2. **Write the Formulas for the Terms:**
The \( n \)-th term of a GP is given by:
\[ T_n = a \cdot r^{n-1} \]
Therefore, we can express the 4th and 6th terms as:
- 4th term: \( T_4 = a \cdot r^{4-1} = a \cdot r^3 = 10 \) (Equation 1)
- 6th term: \( T_6 = a \cdot r^{6-1} = a \cdot r^5 = 40 \) (Equation 2)

3. **Set Up the Equations:**
From Equation 1:
\[ a \cdot r^3 = 10 \tag{1} \]
From Equation 2:
\[ a \cdot r^5 = 40 \tag{2} \]

4. **Divide Equation 2 by Equation 1:**
\[ \frac{a \cdot r^5}{a \cdot r^3} = \frac{40}{10} \]
Simplifying gives:
\[ r^2 = 4 \]
Taking the square root, we find:
\[ r = 2 \quad (\text{since the common ratio is positive}) \]

5. **Substitute \( r \) Back into Equation 1:**
Now substitute \( r = 2 \) into Equation 1 to find \( a \):
\[ a \cdot (2^3) = 10 \]
This simplifies to:
\[ a \cdot 8 = 10 \implies a = \frac{10}{8} = \frac{5}{4} \]

6. **Find the Number of Terms:**
The last term of the GP is given as 640. We can express this as:
\[ T_n = a \cdot r^{n-1} = 640 \]
Substituting the values of \( a \) and \( r \):
\[ \frac{5}{4} \cdot 2^{n-1} = 640 \]
Multiplying both sides by 4:
\[ 5 \cdot 2^{n-1} = 2560 \]
Dividing both sides by 5:
\[ 2^{n-1} = 512 \]
Since \( 512 = 2^9 \), we have:
\[ n - 1 = 9 \implies n = 10 \]

### Final Answers:
- First term \( a = \frac{5}{4} \)
- Common ratio \( r = 2 \)
- Number of terms \( n = 10 \)