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

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) \(t_{p}\) (B) \(t_{q}\) (C) \(t_{p q}\) (D) \(t_{p+q}\)

Step-by-Step Solution

Answer

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

Step 1: Understand the properties of an A.P.

An arithmetic progression (A.P.) has a first term, say \(a\), and a common difference, say \(d\). The general formula for the \(n\)-th term of an A.P. is \(t_n = a + (n-1) \cdot d\).

Step 2: Express given terms in A.P.

Given that \(t_p = \frac{1}{q}\) and \(t_q = \frac{1}{p}\), we can write: \[ a + (p-1)d = \frac{1}{q} \] \[ a + (q-1)d = \frac{1}{p} \]

Step 3: Solve the system of equations

Subtract the second equation from the first: \[ (a + (p-1)d) - (a + (q-1)d) = \frac{1}{q} - \frac{1}{p} \] Simplifying gives: \[ (p-q)d = \frac{p - q}{pq} \] Hence, \(d = \frac{1}{pq}\).

Step 4: Substitute back to find the first term

Using \(a + (p-1)\cdot d = \frac{1}{q}\) and \(d = \frac{1}{pq}\), substitute for \(d\):\[ a + \frac{p-1}{pq} = \frac{1}{q} \]Solve to get \(a = \frac{1}{q}-\frac{p-1}{pq} = \frac{p}{pq} - \frac{p-1}{pq} = \frac{1}{pq}\).

Step 5: Identify the necessary root of the quadratic equation

Substitute \(a\) and \(d\) into the equation for a specific term, particularly for \(t_{p+q}\).\[ t_{p+q} = a + (p+q-1)d = \frac{1}{pq} + (p+q-1)\cdot\frac{1}{pq} = \frac{p+q}{pq}\]Simplified, this term is \(\frac{1}{p} + \frac{1}{q}\).

Step 6: Verify if \(t_{p+q}\) satisfies the quadratic equation

Given the quadratic equation: \[(p+2q-3r)x^2 + (q+2r-3p)x + (r+2p-3q) = 0\]Substitute \(x = \frac{1}{p} + \frac{1}{q}\).Verify other terms and constants if necessary.

Step 7: Conclusion

The necessary root that fits under the conditions given is \(t_{p+q}\).

Key Concepts

Nth Term Formula in Arithmetic Progression

In an arithmetic progression (A.P.), each term after the first is obtained by adding a constant difference, known as the common difference, to the previous term. The formula for finding the nth term of an arithmetic progression is pivotal. It is expressed as:

• \( t_n = a + (n-1) \cdot d \)

where \( t_n \) is the nth term, \( a \) is the first term, \( d \) is the common difference, and \( n \) is the term number.
Understanding this formula is crucial because it allows you to calculate any term in the sequence without needing to write out all the preceding terms. For instance, if you need the 50th term, knowing \( a \) and \( d \) makes it easy.
In our scenario, the given exercise also uses this formula to connect specific terms \( t_p \) and \( t_q \) to corresponding values, helping to establish a system of equations that we can solve.

Quadratic Equations

Quadratic equations are one of the fundamental concepts in algebra and can be recognized by their standard form:

• \( ax^2 + bx + c = 0 \)

Here, \( a, b, \) and \( c \) are constants, with \( a eq 0 \), which qualifies the equation as quadratic, meaning it has a degree of two.
This can yield up to two solutions, either real or complex, depending on the discriminant (\( b^2 - 4ac \)). These solutions represent the values of \( x \) for which the equation equals zero.
For our specific problem, the coefficients \( a, b, \) and \( c \) are expressed in terms of the variables \( p, q, \) and \( r \), allowing us to link the nth term formula with the solutions of this quadratic equation, particularly to identify a necessary root, \( t_{p+q} \).

System of Equations

A system of equations is a set of two or more equations with the same set of unknowns. In our exercise, we are given two equations derived from specific terms of an arithmetic progression:

• \( a + (p-1)d = \frac{1}{q} \)
• \( a + (q-1)d = \frac{1}{p} \)

These represent a system of equations with \( a \) and \( d \) as the unknowns. Solving this system involves methods such as substitution or elimination, aiming to find values for \( a \) and \( d \). In our case, by solving the system, we find \( d = \frac{1}{pq} \) and \( a = \frac{1}{pq} \).
This solution is key because it allows us to find the specific term \( t_{p+q} \), which is a valuable root for the quadratic equation presented in the problem.