Problem 128 - A Complete Resource Book in Mathematics for JEE Main

Question

The number of ways in which three numbers in A.P. can be selected from \(1,2,3, \ldots, n\) is (A) \(\frac{n(n-2)}{4}\), when \(n\) is even (B) \(\frac{1}{4}(n-1)^{2}\), when \(n\) is odd (C) \(\frac{n(n-2)}{2}\), when \(n\) is even (D) None of these

Step-by-Step Solution

Answer

(A) for even n, (B) for odd n.

Step 1: Understanding the Problem

The problem asks us to determine the number of ways to choose three numbers in an arithmetic progression (A.P.) from the numbers 1 through n and to pick the correct expression for even and odd values of n.

Step 2: Identifying Properties of A.P.

In an arithmetic progression, if the first term is a and the common difference is d, the next three terms can be given as a, a+d, a+2d. For these to be in the range from 1 to n, it means: \( a, a+d, a+2d \leq n \). This also translates to \( a + 2d \leq n \).

Step 3: Formulating the Conditions

If a, a+d, a+2d are in A.P., then the condition becomes: \( 1 \leq a \leq n-2d \). For each value of d, we calculate valid values of a.

Step 4: Calculate Total Ways for n Even

For n even, calculate the sum of choices for each d. The number of choices for a is \( n-2d \). Therefore, sum over all possible d values : \( \sum_{d=1}^{\lfloor n/2 \rfloor} (n - 2d) \). This simplifies to \( \frac{n(n-2)}{4} \).

Step 5: Calculate Total Ways for n Odd

For n odd, the calculation follows similarly, but due to the properties of summation and the parity of n, it simplifies to \( \frac{1}{4}(n-1)^2 \).

Step 6: Comparing Options and Deciding

For the choices given: (A) is correct for even n, and (B) for odd n. Verify calculations align with the problem choices. Options correspond to the derived formulae.

Key Concepts

Combinatorics

Combinatorics deals with counting and arranging objects. It's particularly useful in probability and algebra where it helps us analyze different combinations and selections. In our exercise, we want to find the number of ways to select three numbers that form an arithmetic progression (A.P.) from a set of numbers from 1 to n.
An important concept here is the combination, which is a way of selecting items from a larger set such that the order does not matter. For example, we need to find combinations of numbers that form sequences like (a, a+d, a+2d).

• The count depends on factors like n (the total numbers available) and whether n is odd or even.


• Using combinatorics, we'll calculate how many valid (a, d) pairs exist that make sure the selected numbers fit within the range of available numbers.

The calculations differ for even and odd values of n, because of subtle differences in symmetry and pattern completion.

Sequence Selection

Sequence selection involves choosing numbers such that they form a particular pattern, in this case, an arithmetic progression. An arithmetic progression is a sequence of numbers such that the difference between consecutive terms is constant.
For the selection of sequences, we set some constraints based on the definition of arithmetic progressions. Let's break down how these affect our number choices:

• We need the three terms a, a+d, and a+2d to be selected such that they are from the range of 1 to n.


• The condition a + 2d ≤ n ensures that all numbers fit within the given range.


• We also consider each possible value of d, the common difference, and then find suitable starting numbers, a, for each d.

This constraint handling is crucial to efficiently determining valid sequences from a fixed set of numbers.

Mathematical Proofs

Mathematical proofs are arguments that demonstrate why a statement is true. In solving the problem, we needed to derive formulas for different scenarios based on whether n is odd or even.
To prove these formulas, we use a combination of sequence rules and arithmetic:

• For even n, we notice that the sum of valid selections can be simplified to \( \frac{n(n-2)}{4} \).


• This involves summation over possible values of the common difference d and simplifies through mathematical manipulation.


• For odd n, similarly, the proof involves recognizing patterns of summation that simplify to \( \frac{1}{4}(n-1)^2 \).

Each formula is derived by establishing a clear mathematical pathway, starting from conditions of the sequence and simplifying iteratively. This structured approach ensures our derived expressions accurately describe the number of valid sequence selections under the conditions given.