The number of sets of three distinct elemetns that can be chosen from the set `{2^(1),2^(2),2^(3),…….,2^(200)}` such that the three elements form an increasing geometric progression

Options

A. 9900
B. 19900
C. 9950
D. 10000

Answer

To solve the problem of finding the number of sets of three distinct elements from the set `{2^(1), 2^(2), 2^(3), …, 2^(200)}` such that the three elements form an increasing geometric progression (GP), we can follow these steps:

### Step 1: Understand the Set and the Condition for GP
The set consists of powers of 2, specifically:
- \(2^1, 2^2, 2^3, \ldots, 2^{200}\)

For three numbers \(2^a, 2^b, 2^c\) (where \(a < b < c\)) to be in geometric progression, they must satisfy the condition:
\[ 2b = a + c \]
This means that the middle term \(2^b\) must be the geometric mean of the first and last terms \(2^a\) and \(2^c\).

### Step 2: Express the Condition in Terms of Indices
From the GP condition:
\[ 2b = a + c \]
We can rearrange this to find \(b\):
\[ b = \frac{a + c}{2} \]
This implies that \(a + c\) must be even, which means \(a\) and \(c\) must either both be odd or both be even.

### Step 3: Count Odd and Even Indices
The indices range from 1 to 200.
- Odd indices: \(1, 3, 5, \ldots, 199\) (Total: 100 odd numbers)
- Even indices: \(2, 4, 6, \ldots, 200\) (Total: 100 even numbers)

### Step 4: Choose Indices for GP
To form a valid GP:
1. **Choose two odd indices \(a\) and \(c\)**:
- The number of ways to choose 2 odd indices from 100 is given by:
\[ \binom{100}{2} \]

2. **Choose two even indices \(a\) and \(c\)**:
- The number of ways to choose 2 even indices from 100 is also:
\[ \binom{100}{2} \]

### Step 5: Calculate Total Combinations
The total number of sets of three distinct elements that can be chosen is the sum of the combinations from odd and even indices:
\[ \text{Total} = \binom{100}{2} + \binom{100}{2} \]
Calculating \(\binom{100}{2}\):
\[ \binom{100}{2} = \frac{100 \times 99}{2} = 4950 \]
Thus, the total number of sets is:
\[ \text{Total} = 4950 + 4950 = 9900 \]

### Final Answer
The number of sets of three distinct elements that can be chosen from the set such that they form an increasing geometric progression is **9900**.

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