If the product of five consecutive terms of G.P. is 1024, then find the third term of this set.

Answer

To solve the problem, we need to find the third term of a geometric progression (G.P.) given that the product of five consecutive terms is 1024. Let's denote the first term of the G.P. as \( a \) and the common ratio as \( r \).

### Step-by-Step Solution:

1. **Identify the Terms of the G.P.**:
The five consecutive terms of the G.P. can be expressed as:
- First term: \( a \)
- Second term: \( ar \)
- Third term: \( ar^2 \)
- Fourth term: \( ar^3 \)
- Fifth term: \( ar^4 \)

2. **Set Up the Product Equation**:
The product of these five terms is given by:
\[ P = a \cdot ar \cdot ar^2 \cdot ar^3 \cdot ar^4 \]
This simplifies to:
\[ P = a^5 \cdot r^{0+1+2+3+4} = a^5 \cdot r^{10} \]
We know that this product equals 1024:
\[ a^5 \cdot r^{10} = 1024 \]

3. **Express 1024 as a Power of 2**:
We can express 1024 as a power of 2:
\[ 1024 = 2^{10} \]
Therefore, we have:
\[ a^5 \cdot r^{10} = 2^{10} \]

4. **Equate the Powers**:
We can rewrite \( a^5 \cdot r^{10} \) in a different form:
\[ (ar^2)^5 = a^5 \cdot r^{10} \]
This means:
\[ ar^2 = 2^2 = 4 \]
Thus, the third term \( T_3 \) of the G.P. is:
\[ T_3 = ar^2 = 4 \]

5. **Conclusion**:
The third term of the set is:
\[ \boxed{4} \]