To solve the problem of finding the number of ways to select 3 distinct numbers from the set \(\{3^1, 3^2, 3^3, \ldots, 3^{101}\}\) such that they form a geometric progression (G.P.), we can follow these steps:
### Step 1: Understanding the Condition for G.P.
In a geometric progression, the middle term squared is equal to the product of the other two terms. If we denote the selected terms as \(3^A\), \(3^B\), and \(3^C\) (where \(A\), \(B\), and \(C\) are distinct integers), the condition for these to be in G.P. is:
\[
(3^B)^2 = 3^A \cdot 3^C
\]
This simplifies to:
\[
3^{2B} = 3^{A+C}
\]
Thus, we have:
\[
2B = A + C
\]
### Step 2: Relating to Arithmetic Progression
The equation \(2B = A + C\) implies that \(A\), \(B\), and \(C\) are in arithmetic progression (A.P.). Therefore, we need to select \(A\), \(B\), and \(C\) such that they satisfy this condition.
### Step 3: Selecting Values for A, B, and C
The values of \(A\), \(B\), and \(C\) must be distinct integers from the set \(\{1, 2, 3, \ldots, 101\}\).
1. **Finding the Range**:
- The smallest value for \(A\) can be 1, and the largest value for \(C\) can be 101.
- Since \(B\) is the middle term, we can express \(A\) and \(C\) in terms of \(B\):
- Let \(A = B - d\) and \(C = B + d\) for some integer \(d\).
2. **Determining Valid Values**:
- For \(A\) to be at least 1, we must have:
\[
B - d \geq 1 \implies d \leq B - 1
\]
- For \(C\) to be at most 101, we must have:
\[
B + d \leq 101 \implies d \leq 101 - B
\]
- Therefore, \(d\) must satisfy:
\[
d \leq \min(B - 1, 101 - B)
\]
### Step 4: Counting the Valid Combinations
Now we can count the number of valid combinations for each possible value of \(B\):
- The possible values for \(B\) range from 2 to 100 (since \(B\) cannot be 1 or 101 to ensure \(A\) and \(C\) are distinct).
- For each \(B\), the maximum value of \(d\) is:
\[
d_{\text{max}} = \min(B - 1, 101 - B)
\]
- The number of valid \(d\) values for each \(B\) is \(d_{\text{max}}\).
### Step 5: Summing Over All Possible B
We now sum \(d_{\text{max}}\) for each \(B\) from 2 to 100:
\[
\text{Total Ways} = \sum_{B=2}^{100} \min(B - 1, 101 - B)
\]
- For \(B = 2\) to \(B = 50\), \(d_{\text{max}} = B - 1\).
- For \(B = 51\), \(d_{\text{max}} = 50\).
- For \(B = 52\) to \(B = 100\), \(d_{\text{max}} = 101 - B\).
Calculating these:
- For \(B = 2\) to \(B = 50\):
\[
\sum_{B=2}^{50} (B - 1) = 1 + 2 + \ldots + 49 = \frac{49 \cdot 50}{2} = 1225
\]
- For \(B = 51\):
\[
d_{\text{max}} = 50
\]
- For \(B = 52\) to \(B = 100\):
\[
\sum_{B=52}^{100} (101 - B) = 49 + 48 + \ldots + 1 = \frac{49 \cdot 50}{2} = 1225
\]
### Final Calculation
Adding these together:
\[
\text{Total Ways} = 1225 + 50 + 1225 = 2500
\]
Thus, the total number of ways to select 3 distinct numbers from the set such that they form a G.P. is **2500**.
