Find the number of terms in the progression `4, 2, 1, …,(1)/(128).`

Explanation

To find the number of terms in the geometric progression (GP) given by \(4, 2, 1, \ldots, \frac{1}{128}\), we can follow these steps:

### Step 1: Identify the first term and common ratio
The first term \(A\) of the GP is \(4\). The common ratio \(R\) can be found by dividing the second term by the first term:
\[ R = \frac{2}{4} = \frac{1}{2} \]

### Step 2: Write the formula for the nth term of a GP
The nth term \(A_n\) of a GP can be expressed using the formula:
\[ A_n = A \cdot R^{n-1} \]
where \(A\) is the first term, \(R\) is the common ratio, and \(n\) is the term number.

### Step 3: Set up the equation for the last term
We know that the last term \(A_n\) is \(\frac{1}{128}\). Therefore, we can set up the equation:
\[ \frac{1}{128} = 4 \cdot \left(\frac{1}{2}\right)^{n-1} \]

### Step 4: Simplify the equation
First, we can express \(\frac{1}{128}\) in terms of powers of \(2\):
\[ \frac{1}{128} = \frac{1}{2^7} \]
Now, substituting this into the equation gives us:
\[ \frac{1}{2^7} = 4 \cdot \left(\frac{1}{2}\right)^{n-1} \]
We can express \(4\) as \(2^2\):
\[ \frac{1}{2^7} = 2^2 \cdot \left(\frac{1}{2}\right)^{n-1} \]
This simplifies to:
\[ \frac{1}{2^7} = \frac{2^2}{2^{n-1}} = \frac{1}{2^{n-1-2}} = \frac{1}{2^{n-3}} \]

### Step 5: Set the exponents equal to each other
Since the bases are the same, we can equate the exponents:
\[ -7 = -(n-3) \]
This simplifies to:
\[ -7 = -n + 3 \]
Rearranging gives:
\[ n - 3 = 7 \implies n = 10 \]

### Conclusion
Thus, the number of terms in the given progression is \(10\).

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