Problem 128 - A Complete Resource Book in Mathematics for JEE Main 2017

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 \(n\) and (B) for odd \(n\).

Step 1: Understand the Arithmetic Progression (A.P.)

In an arithmetic progression (A.P.), the numbers have a constant difference between consecutive terms. If we choose numbers \(a-d\), \(a\), and \(a+d\) from a set \(1, 2, 3, \, \ldots, n\), then \(a-d\) must be the smallest and \(a+d\) the largest of the three terms.

Step 2: Determine Conditions for Numbers in A.P.

For the three numbers \(a-d\), \(a\), and \(a+d\) to all be selected from \(\{1, 2, 3, \ldots, n\}\), the following must hold: \(a-d \geq 1\) and \(a+d \leq n\). Simplifying these, we get \(d < a\) and \(a+d \leq n\), which leads to \(a \leq n-d\). Therefore, the possible values for \(a\) are constrained by the value of \(d\).

Step 3: Calculate the Number of Ways for Each d

For each \(d\), \(a\) can range from \(d+1\) to \(n-d\). Therefore, the number of values for \(a\) is \(n-2d\).

Step 4: Total Number of Ways for all Possible d

\(d\) can range from 1 to \(\left\lfloor \frac{n-1}{2} \right\rfloor\) because for \(a \geq d+1\), \(d\) must be less than \(\frac{n}{2}\). Summing over all possible values of \(d\), we get: \(\sum_{d=1}^{\left\lfloor \frac{n-1}{2} \right\rfloor} (n-2d)\). This summation will give the total number of ways to choose \(a\), \(a+d\), and \(a-d\).

Step 5: Solve the Summation

Let's calculate \(\sum_{d=1}^{k} (n-2d)\), where \(k = \left\lfloor \frac{n-1}{2} \right\rfloor\). The summation is: \[ n\cdot k - 2\cdot \sum_{d=1}^{k} d \] The arithmetic series sum \(\sum_{d=1}^{k} d\) is \(\frac{k(k+1)}{2}\). Therefore, the total number of ways is:\[ n\cdot k - k(k+1) = k(n-k-1) \].

Step 6: Analyze Based on n being Even or Odd

For \(n\) even, \(k = \frac{n}{2}-1\). Therefore,\[ \frac{n}{2} \left(n - \frac{n}{2}\right) - \frac{n}{2} = \frac{n(n-2)}{4} \]For \(n\) odd, \(k = \frac{n-1}{2}\), thus:\[ \frac{1}{4}(n-1)^2 \].

Step 7: Conclusion

Compare the derived formulas for different \(n\) with the given options.- For even \(n\), our formula \(\frac{n(n-2)}{4}\) matches option (A).- For odd \(n\), our formula \(\frac{1}{4}(n-1)^2\) matches option (B). Thus, the correct answer is (A) for even \(n\) and (B) for odd \(n\).

Key Concepts

combinatorics

Combinatorics is a branch of mathematics focused on counting, arrangement, and combination of objects within a set, according to specified rules. In this particular problem, combinatorics is used to find the number of different ways to select three numbers in arithmetic progression from a given integer sequence. Combinatorics often involves calculating probabilities, combinations, permutations, and more practical, everyday applications like arranging items or organizing teams.

In this case, the challenge is determining how many sets of numbers can be chosen such that they form an arithmetic progression (A.P.). When we speak of A.P., we imply a sequence where the difference between consecutive elements remains constant, denoted as "d" here. Counting such sequences entails understanding constraints on the selection based on the given range of numbers, making combinatorics crucial to derive the correct formula by summing over possible differences 'd' within the sequence.

series summation

Series summation is an essential tool in mathematics, particularly when dealing with sequences and their sums. The sum of a series is calculated by adding all elements of a sequence according to specific rules or formulas. In many scenarios, formulas for series summation help simplify what would otherwise be lengthy calculations.

In the solution presented for this exercise, series summation comes into play when computing the total number of ways to choose three numbers in A.P. for every possible difference, 'd'. The formula \(\sum_{d=1}^{k} (n-2d)\) is established to ascertain this sum, where 'k' is the upper bound for differences. To compute this, we utilize the arithmetic series sum \(\sum_{d=1}^{k} d\), which is given by \(\frac{k(k+1)}{2}\). This simplification is pivotal in obtaining results quickly and efficiently. It underpins understanding how sequences accumulate value when parts are successively added.

integer sequences

Integer sequences are ordered lists of integers that follow a specific pattern or rule guiding their progression. They appear frequently in mathematical exercises, often necessitating identification of the pattern, computation of terms, or summation thereof, as seen in this exercise.

Here, we deal with the sequence of integers starting from 1 up to a given number 'n'. The main focus is selecting subsequences of three numbers that form an arithmetic progression (A.P.). Integer sequences lend themselves to a variety of problems, such as figuring out terms and relationships between them.

Understanding integer sequences includes recognizing their behavior when subjected to operations like sum or transformation, such as shifting by a constant 'd' to form terms of the A.P. This understanding makes it possible to apply mathematical tools and derive precise results, such as the number of ways a specified subsequence can be formed from the larger integer sequence.