If the nth terms of the progression 5, 10, 20, … and progression 1280, 640, 320,… are equal, then find the value of n.

Explanation

To solve the problem, we need to find the value of \( n \) such that the \( n \)th terms of the two given sequences are equal.

1. **Identify the sequences:**
- The first sequence is \( 5, 10, 20, \ldots \)
- The second sequence is \( 1280, 640, 320, \ldots \)

2. **Determine the type of sequences:**
- The first sequence is a geometric progression (GP) where the first term \( a = 5 \) and the common ratio \( r = 2 \).
- The second sequence is also a geometric progression where the first term \( a = 1280 \) and the common ratio \( r = \frac{1}{2} \).

3. **Write the formula for the \( n \)th term of each sequence:**
- For the first sequence (GP):
\[ T_n = a \cdot r^{n-1} = 5 \cdot 2^{n-1} \]
- For the second sequence (GP):
\[ T_n = a \cdot r^{n-1} = 1280 \cdot \left(\frac{1}{2}\right)^{n-1} \]

4. **Set the \( n \)th terms equal to each other:**
\[ 5 \cdot 2^{n-1} = 1280 \cdot \left(\frac{1}{2}\right)^{n-1} \]

5. **Simplify the equation:**
- Rewrite \( \left(\frac{1}{2}\right)^{n-1} \) as \( 2^{-(n-1)} \):
\[ 5 \cdot 2^{n-1} = 1280 \cdot 2^{-(n-1)} \]

6. **Multiply both sides by \( 2^{n-1} \) to eliminate the fraction:**
\[ 5 \cdot (2^{n-1})^2 = 1280 \]
\[ 5 \cdot 2^{2(n-1)} = 1280 \]

7. **Divide both sides by 5:**
\[ 2^{2(n-1)} = \frac{1280}{5} \]
\[ 2^{2(n-1)} = 256 \]

8. **Express \( 256 \) as a power of \( 2 \):**
\[ 256 = 2^8 \]

9. **Set the exponents equal to each other:**
\[ 2(n-1) = 8 \]

10. **Solve for \( n \):**
\[ n-1 = 4 \implies n = 5 \]

Thus, the value of \( n \) is \( 5 \).