Question

Let \(T_{r}\) be the \(r\) th term of an A.P. for \(r=1,2,3, \ldots\) If for some positive integers \(m, n\) we have \(T_{m}=\frac{1}{n}\) and \(T_{n}=\frac{1}{m}\) then \(T_{\text {mu }}\) equals : (a) \(\frac{1}{m n}\) (b) \(\frac{1}{m}+\frac{1}{n}\) (c) 1 (d) 0

Short answer

Question: In an A.P., T_m = \frac{1}{n} and T_n = \frac{1}{m}, then the value of the term T_{\text{mu}} is: a) \frac{1}{mn} b) \frac{1}{m} + \frac{1}{n} c) 1 d) 0 Answer: None of the given options are correct. The correct value of T_{\text{mu}} is \frac{mn - n - m + 1}{n}.

Step-by-step solution

Recall the formula for finding a term in an A.P.

The formula for finding the r-th term (T_r) of an arithmetic progression is given by: T_r = a + (r - 1)d where 'a' is the first term, 'd' is the common difference, and 'r' is the term number.

Substitute the given conditions

We have T_m = \frac{1}{n} and T_n = \frac{1}{m}. Let's substitute these values in the formula: For T_m: a + (m - 1)d = \frac{1}{n} For T_n: a + (n - 1)d = \frac{1}{m}

Solve for 'a' and 'd'

We now have a system of two linear equations with two variables (a and d). Let's solve this system: Subtract the first equation from the second equation: ((n - 1)d - (m - 1)d) = \frac{1}{m} - \frac{1}{n} Solve for d: (n-m)d = \frac{n-m}{mn} Now, divide both sides by (n-m): d = \frac{1}{mn} Plug the value of d into any of the equations to find the value of 'a'. Let's use the first equation: a + (m-1)\frac{1}{mn} = \frac{1}{n} Solve for a: a = \frac{1}{n} - \frac{m-1}{mn} = \frac{1}{n}(1 - \frac{m-1}{m}) Now, we have the values for 'a' and 'd' in terms of 'm' and 'n'.

Substitute 'a' and 'd' into the formula while solving for T_mu

mu = m * n since we're looking for T_{\text{mu}}) Now substitute the values of 'a' and 'd' into the formula for T_{\text{mu}}. T_{\text{mu}} = a + (\text{mu} - 1)d T_{\text{mu}} = \frac{1}{n}(1 - \frac{m-1}{m}) + (mn - 1) \frac{1}{mn} Simplify the expression: T_{\text{mu}} = \frac{1}{n} - \frac{1}{n^2} + \frac{mn - 1}{mn} Now, simplify and combine the fractions: T_{\text{mu}} = \frac{mn^2 - n^2 + mn^3 - mn^2}{n^2mn} = \frac{mn^2(mn - n - m + 1)}{n^2mn} The mn terms in the numerator and denominator cancel out, resulting in the following expression: T_{\text{mu}} = \frac{n(mn - n - m + 1)}{n^2} Now, divide through by n: T_{\text{mu}} = \frac{mn - n - m + 1}{n}

Identify the correct option

Now we can compare our solution for T_{\text{mu}} with the given options: (a) \frac{1}{mn} - Incorrect, our solution is not in this form. (b) \frac{1}{m} + \frac{1}{n} - Incorrect, our solution is different. (c) 1 - Incorrect, our solution is not equal to 1. (d) 0 - Incorrect, our solution is not equal to 0. The correct answer is not given among the options. The value of T_{\text{mu}} is equal to \frac{mn - n - m + 1}{n}.

Key concepts

AP Formula

Understanding the arithmetic progression (AP) formula is essential to finding any term in a sequence. It's given by the expression
\( T_r = a + (r - 1)d \)
, where \( T_r \) represents the term you're looking for, \( a \) is the first term of the sequence, \( d \) is the common difference between the terms, and \( r \) is the term number. This formula is incredibly useful because it establishes a clear linear relationship between the terms and their position in the sequence. Instead of adding the common difference repeatedly, you can simply plug in the term number \( r \) and calculate its value.

Term of an AP

The term of an AP can be thought of as an individual member of the arithmetic progression. When we discuss an AP, we're talking about a sequence of numbers where each term after the first is obtained by adding a constant fixed amount known as the common difference (
\( d \)
) to the previous one.
To find an arbitrary term like \( T_{m} \) or \( T_{n} \) in our example, we modify the AP formula by substituting the term number for \( r \). This gets us to precise values, showing the term's position in the sequence and its relationship to other terms.

Linear Equations

In our step-by-step solution, we encounter linear equations, which are fundamental in solving problems involving an AP. A linear equation is an algebraic equation in which each term is either a constant or the product of a constant and a single variable. When we have expressions like \( T_{m} = \frac{1}{n} \) and \( T_{n} = \frac{1}{m} \), and substitute them into the AP formula, we end up with two linear equations.
These equations form a system that we can solve using standard algebraic techniques like substitution, elimination, or graphic methods. By solving for the unknowns, which in our case are the first term (\( a \)) and the common difference (\( d \)), we can then calculate specific terms in the sequence, such as \( T_{\text{mu}} \). This demonstrates the interconnectedness of algebraic concepts and highlights the importance of mastering linear equations to handle various mathematical problems.