The 3rd and 6th terms of a G.P. are 40 and 320, then find the progression.

Explanation

To find the geometric progression (G.P.) given that the 3rd and 6th terms are 40 and 320 respectively, we can follow these steps:

### Step-by-Step Solution:

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

2. **Set up the equations for the 3rd and 6th terms**:
From the problem, we have:
- The 3rd term \( T_3 = ar^{3-1} = ar^2 = 40 \)
- The 6th term \( T_6 = ar^{6-1} = ar^5 = 320 \)

This gives us two equations:
\[ ar^2 = 40 \quad \text{(1)} \]
\[ ar^5 = 320 \quad \text{(2)} \]

3. **Divide the two equations**:
We can divide equation (2) by equation (1) to eliminate \( a \):
\[ \frac{ar^5}{ar^2} = \frac{320}{40} \]
Simplifying this gives:
\[ r^{5-2} = \frac{320}{40} \]
\[ r^3 = 8 \]

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

5. **Substitute \( r \) back to find \( a \)**:
Now that we have \( r \), we can substitute it back into equation (1) to find \( a \):
\[ ar^2 = 40 \]
\[ a(2^2) = 40 \]
\[ 4a = 40 \]
\[ a = 10 \]

6. **Write the G.P.**:
Now that we have both \( a \) and \( r \), we can write the terms of the G.P.:
- First term: \( a = 10 \)
- Second term: \( ar = 10 \times 2 = 20 \)
- Third term: \( ar^2 = 10 \times 2^2 = 40 \)
- Fourth term: \( ar^3 = 10 \times 2^3 = 80 \)
- Fifth term: \( ar^4 = 10 \times 2^4 = 160 \)
- Sixth term: \( ar^5 = 10 \times 2^5 = 320 \)

Thus, the progression is:
\[ 10, 20, 40, 80, 160, 320 \]

### Final Answer:
The geometric progression is \( 10, 20, 40, 80, 160, 320 \).

---