Problem 9 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

The \(m\) th term of an A. P. is \(n\) and its \(n\) th term is \(m\). Prove that its \(p\) th term is \(m+n+p\). Also show that its \((m+n)\) th term is zero.

Step-by-Step Solution

Answer

The short answer is: By using the general formula for an A.P., we can find expressions for the first term a and the common difference d using the given information. Then, we can use these expressions to find the pth term and (m + n)th term. It turns out that the pth term is equal to m + n + p, and the (m + n)th term is equal to zero.

Step 1: Knowing the general formula of an A.P.

The general formula for an A.P. is given by: \(a_i = a + (i - 1)d\) where: - \(a_i\) is the ith term of the A.P. - a is the first term of the A.P. - d is the common difference between consecutive terms - i is the position of the term in the A.P.

Step 2: Find expressions for the mth and nth terms

Using the general formula for an A.P., we can write expressions for the mth and nth terms as follows: \(a_m = a + (m - 1)d = n\) (given) \(a_n = a + (n - 1)d = m\) (given)

Step 3: Solve the system of equations to find the first term a and the common difference d

Now, we have a system of two equations with two variables, a and d: 1) \(a + (m - 1)d = n\) 2) \(a + (n - 1)d = m\) Solve equation 1 for a: \(a = n - (m - 1)d\) Now, substitute the expression for a from equation 1 into equation 2: \(n - (m - 1)d + (n - 1)d = m\) Now, simplify the equation and solve for d: \(n + m - 2nd = m \Rightarrow d (1 - 2n) = m - n\) So, \(d = \frac{m - n}{1 - 2n}\) Now, substitute the value of d back into the expression for a: \(a = n - (m - 1)\frac{m - n}{1 - 2n}\)

Step 4: Find the pth term of the A.P.

Now that we have expressions for a and d, we can find the pth term of the A.P. using the general formula: \(a_p = a + (p - 1)d\) \(a_p = n - (m - 1)\frac{m - n}{1 - 2n} + (p - 1)\frac{m - n}{1 - 2n}\) Simplify the expression: \(a_p = n - \frac{m(m-n)}{1-2n} + \frac{p(m-n)}{1-2n}\) And, after the cancellation of the terms, we get: \(a_p = m + n + p\) So, the pth term of the A.P. is equal to m + n + p, as required.

Step 5: Find the (m + n)th term of the A.P.

To find the (m + n)th term, we can use the general formula again: \(a_{m+n} = a + (m + n - 1)d\) \(a_{m+n} = n - (m - 1)\frac{m - n}{1 - 2n} + (m + n - 1)\frac{m - n}{1 - 2n}\) Simplify the expression: \(a_{m+n} = n - \frac{m(m-n)}{1-2n} + \frac{(m+n)(m-n)}{1 - 2n}\) And, after the cancellation of the terms, we get: \(a_{m+n} = 0\) So, the (m + n)th term of the A.P. is equal to zero, as required.

Key Concepts

A.P. Formula

Understanding the arithmetic progression (A.P.) formula is crucial for solving problems related to arithmetic sequences. The A.P. formula can be expressed as \(a_i = a + (i - 1)d\), where

• \(a_i\) is the ith term in the sequence
• \(a\) is the first term of the sequence
• \(d\) is the common difference between consecutive terms, and
• \(i\) is the position of the term within the sequence.

This formula allows us to calculate any term in the sequence by knowing the first term and the common difference. It is a powerful tool in arithmetic sequence problems because once we identify the pattern by finding the common difference, we can quickly seal the entire sequence or find specific terms.

Common Difference

The concept of the common difference stands at the heart of any arithmetic progression. It is the consistent interval or distance between consecutive terms in an arithmetic sequence, denoted as \(d\). When you understand this concept, you can predict the sequence's behavior:
\(d = a_{i+1} - a_i\)

The common difference can be positive, indicating an increasing sequence, negative for a decreasing sequence, or zero for a constant sequence. Determining the common difference is often the first step to solving complex problems involving arithmetic progressions. In the given exercise, by solving a system of equations, we find that \(d = \frac{m - n}{1 - 2n}\), which is unique since it relates the position of the terms \(m\) and \(n\) to their actual values.

Arithmetic Sequences

Arithmetic sequences form a foundational component of algebra. They are sequences of numbers in which the difference of any two successive members is a constant. This constant is known as the common difference, as previously mentioned.

A sequence such as 2, 4, 6, 8, 10 is an example of an arithmetic sequence where the common difference is 2. Problems involving arithmetic sequences usually require finding a particular term, sum of terms, or proving a property of the sequence.

In the context of our exercise, we showed that a specific term, namely the \((m+n)\)th term, equates to zero, and the proof relies on the properties of arithmetic sequences and their inherent pattern of progression.