Question

The number of ways of selecting 3 numbers that are in Arithmetic Progression (A.P.) from the set \(\{1, 2, 3, \ldots, 100\}\) is:

• A. 1600
• B. 1650
• C. 2450
• D. 1667

Hint:

The number of 3-term arithmetic progressions in a set \(\{1, 2, \ldots, n\}\) is \(\sum_{d=1}^{\lfloor (n-1)/2 \rfloor} (n - 2d)\).

Answer 1

The Correct Option is C

Let the three numbers in arithmetic progression be \(a, a + d, a + 2d\), where \(a\) is the first term and \(d>0\) is the common difference. Since all numbers must lie in the set \(\{1, 2, \ldots, 100\}\), we have the constraint: \[ 1 \leq a \leq 100 - 2d \] because the largest term \(a + 2d\) must be at most 100. For each fixed \(d\), the number of valid choices for \(a\) is: \[ 100 - 2d \] The common difference \(d\) can take integer values from 1 up to: \[ d \leq \frac{100 - 1}{2} = 49.5 \implies d = 1, 2, \ldots, 49 \] Therefore, total number of 3-term arithmetic progressions is: \[ \sum_{d=1}^{49} (100 - 2d) = 49 \times 100 - 2 \times \frac{49 \times 50}{2} = 4900 - 2450 = 2450 \]