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 progressions are equal.

### Step 1: Identify the type of progressions
The first progression is \( 5, 10, 20, \ldots \). This is a geometric progression (GP) with:
- First term \( a_1 = 5 \)
- Common ratio \( r_1 = \frac{10}{5} = 2 \)

The second progression is \( 1280, 640, 320, \ldots \). This is also a GP with:
- First term \( a_2 = 1280 \)
- Common ratio \( r_2 = \frac{640}{1280} = \frac{1}{2} \)

### Step 2: Write the formula for the \( n \)-th term of each GP
The \( n \)-th term of a geometric progression can be calculated using the formula:
\[ A_n = a \cdot r^{n-1} \]

For the first progression:
\[ A_n = 5 \cdot 2^{n-1} \]

For the second progression:
\[ B_n = 1280 \cdot \left(\frac{1}{2}\right)^{n-1} \]

### Step 3: Set the \( n \)-th terms equal to each other
Since we need to find \( n \) such that \( A_n = B_n \):
\[ 5 \cdot 2^{n-1} = 1280 \cdot \left(\frac{1}{2}\right)^{n-1} \]

### Step 4: Simplify the equation
We can rewrite \( \left(\frac{1}{2}\right)^{n-1} \) as \( \frac{1}{2^{n-1}} \):
\[ 5 \cdot 2^{n-1} = 1280 \cdot \frac{1}{2^{n-1}} \]

Multiplying both sides by \( 2^{n-1} \) gives:
\[ 5 \cdot (2^{n-1})^2 = 1280 \]
\[ 5 \cdot 2^{2(n-1)} = 1280 \]

### Step 5: Divide both sides by 5
\[ 2^{2(n-1)} = \frac{1280}{5} \]
Calculating \( \frac{1280}{5} \):
\[ \frac{1280}{5} = 256 \]
Thus, we have:
\[ 2^{2(n-1)} = 256 \]

### Step 6: Express 256 as a power of 2
We know that:
\[ 256 = 2^8 \]
So we can write:
\[ 2^{2(n-1)} = 2^8 \]

### Step 7: Set the exponents equal to each other
Since the bases are the same, we can set the exponents equal:
\[ 2(n-1) = 8 \]

### Step 8: Solve for \( n \)
Dividing both sides by 2:
\[ n - 1 = 4 \]
Adding 1 to both sides:
\[ n = 5 \]

### Final Answer
The value of \( n \) is \( 5 \).

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