Problem 110 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

If \(t_{n}\) denotes the \(n\)th term of an A.P. and \(t_{p}=\frac{1}{q}\) and \(t_{q}\) \(=\frac{1}{p}\), then which of the following is necessarily a root of the equation \((p+2 q-3 r) x^{2}+(q+2 r-3 p) x+(r\) \(+2 p-3 q)=0\) (A) \(\bar{t}_{p}\) (B) \(t_{q}\) (C) \(t_{p q}\) (D) \(t_{p+q}\)

Step-by-Step Solution

Answer

\(t_{p+q}\) is the root of the equation.

Step 1: Understanding the Problem

We are given that in an arithmetic progression (A.P.), the nth term is denoted by \(t_n\). We are also given two terms: \(t_p = \frac{1}{q}\) and \(t_q = \frac{1}{p}\). We need to find which option is a root of the provided equation.

Step 2: General Formula for nth Term

The general formula for the nth term of an A.P. is \(t_n = a + (n-1) d\), where \(a\) is the first term and \(d\) is the common difference. Using this formula, we will set up two equations based on the given terms: \(t_p = \frac{1}{q}\) and \(t_q = \frac{1}{p}\).

Step 3: Create Equations

For \(t_p = \frac{1}{q}\): \[ a + (p-1) d = \frac{1}{q} \] For \(t_q = \frac{1}{p}\): \[ a + (q-1) d = \frac{1}{p} \]

Step 4: Solve Simultaneous Equations

Subtract the second equation from the first: \[ (p-q) d = \frac{1}{q} - \frac{1}{p} \] Which simplifies to: \[ (p-q) d = \frac{p-q}{pq} \] \[ d = \frac{1}{pq} \]

Step 5: Find Specific Term

We are asked to consider various \(t_n\) terms as potential roots, such as \(t_p, t_q, t_{pq}\), and \(t_{p+q}\). Using the formula \(t_n = a + (n-1) d\), substitute to find each of these terms' expressions.

Step 6: Substitute Specific Term for Root Testing

For \(t_{p+q}\): \[ t_{p+q} = a + ((p+q)-1)d \] \[ t_{p+q} = \frac{1}{2} \left( \frac{1}{p} + \frac{1}{q} \right) + (p+q-1) \cdot \frac{1}{pq} \] Using the simplified expression for \(d\), substitute \(t_{p+q}\) in given equation and check if it reduces to zero.

Step 7: Check Solution Validity

Calculating, \(t_{p+q} = \frac{1}{p} + \frac{1}{q} + \frac{p+q-1}{pq} = \frac{p+q}{2pq}\). Substitute this into the quadratic equation to verify it results in zero, confirming it as a root.

Key Concepts

nth term

In an arithmetic progression (A.P.), the "nth term" refers to a specific position in the sequence. Every A.P. is characterized by its first term and the constant difference between consecutive terms. You can calculate the value of any term by using the formula:\[ t_n = a + (n-1)d \]where:

• \( t_n \) is the nth term,
• \( a \) is the first term of the sequence,
• \( d \) is the common difference between terms,
• \( n \) is the term number you're interested in.

This formula allows you to find any term in the sequence, assuming you know the first term and the common difference. For example, if the first term of an A.P. is 2, and the common difference is 3, the 5th term would be calculated as follows:\[ t_5 = 2 + (5-1) \cdot 3 = 2 + 12 = 14 \]This concept is fundamental in analyzing sequences because it provides a direct method to calculate any term from just two initial parameters.

common difference

The "common difference" is a key element in an arithmetic progression. It represents the constant difference between any two successive terms in the sequence. When you have a series like 3, 7, 11, 15, etc., the common difference is 4, because:\[ t_{n+1} - t_n = d \]For our series, each term increases by 4. Mathematically, it can be shown that:\[ d = t_2 - t_1 \]or generally:\[ d = a_{n+1} - a_n \]The arithmetic property of having the same difference makes calculations consistent and predictable. Knowing the common difference allows anyone to extend the sequence both forward and backward. You can even derive the entire formula for the nth term by applying the common difference repeatedly from the first term.

simultaneous equations

Simultaneous equations involve finding common solutions to two or more equations with multiple variables. When dealing with arithmetic progressions, simultaneous equations often arise when you need to solve for unknowns like the first term or common difference. Suppose we have two equations representing terms of an A.P.:\[ a + (p-1)d = \frac{1}{q} \]\[ a + (q-1)d = \frac{1}{p} \]To find the common difference \(d\), we can solve these equations together. We typically subtract one equation from the other to eliminate one variable and solve for the remaining one. This results in:\[ (p-q)d = \frac{1}{q} - \frac{1}{p} \]This simplifies to give the value of \(d\), and by substituting back into either original equation, we solve for \(a\). This process illustrates how simultaneous equations help decipher values in related scenarios.

roots of quadratic equation

The "roots of a quadratic equation" are the values of the variable that make the equation equal to zero. Quadratic equations are in the form:\[ ax^2 + bx + c = 0 \]In solving these, you find the roots using the quadratic formula:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]In our problem, you needed to determine the possible roots by substituting term expressions into this quadratic equation:\[(p+2q-3r)x^2 + (q+2r-3p)x + (r+2p-3q) = 0\]By substituting the calculated term, \(t_{p+q}\), and verifying that it results in the equation resolving to zero, it confirms this term is indeed a root. Finding roots helps in understanding the points where the values transform from positive to negative or vice versa, marking important changes.