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Chapter 9: Problem 42

Let \(T_{\mathrm{r}}\) be the rth term of an A.P. whose first term is a and common difference is \(d\). If for some positive integers \(m, n, m \neq n, T_{m}=\frac{1}{n}\) and \(T_{n}=\frac{1}{m}\), then \(a-d\) equals [2004] (a) \(\frac{1}{m}+\frac{1}{n}\) (b) 1 (c) \(\frac{1}{m n}\) (d) 0

Short Answer

Expert verified
\(a-d = \frac{1}{m} + \frac{1}{n}\), which is option (a).

Step by step solution

01

Write the formula for the rth term of an A.P.

The rth term of an arithmetic progression (A.P.) is given by the formula: \( T_r = a + (r-1) \cdot d \), where \( a \) is the first term and \( d \) is the common difference.
02

Express given conditions using the formula

According to the problem, we have two equations: \( T_m = \frac{1}{n} \) and \( T_n = \frac{1}{m} \).\[ T_m = a + (m-1) \cdot d = \frac{1}{n} \]\[ T_n = a + (n-1) \cdot d = \frac{1}{m} \]
03

Subtract the equations to find d

Subtract the first equation from the second:\[ (a + (n-1) \cdot d) - (a + (m-1) \cdot d) = \frac{1}{m} - \frac{1}{n} \]Simplifying gives:\[ (n-m) \cdot d = \frac{1}{m} - \frac{1}{n} \]\[ d = \frac{\frac{1}{m} - \frac{1}{n}}{n-m} \]
04

Calculate a - d

Using the expression for \( d \) derived above and the equation \( a = T_m - (m-1) \cdot d = \frac{1}{n} - (m-1) \cdot d \), solve for \( a - d \).Additionally, we determine:\[ a - d = \frac{1}{m} - (n-1) \cdot d = \frac{1}{m} - \frac{1}{n} \cdot \frac{1}{(n-m)} \times (n-m) = \frac{1}{m} + \frac{1}{n} \]
05

Verify result matches given options

The expression for \( a - d \) we found is \( \frac{1}{m} + \frac{1}{n} \), which corresponds to option (a) in the problem statement.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

rth term formula
An arithmetic progression (A.P.) is a sequence of numbers in which the difference of any two consecutive numbers is always the same. This difference is known as the common difference, denoted by \( d \). One of the most important formulas in an arithmetic progression is the formula for the rth term. This formula allows you to find any specific term in the sequence without listing all the terms.Let's understand the formula: The rth term, denoted as \( T_r \), is given by:\[T_r = a + (r-1) \, \cdot \, d\]In this formula:
  • \( T_r \) is the rth term you want to find.
  • \( a \) is the first term of the sequence.
  • \( r \) is the position of the term you want to find, also known as the 'term number'.
  • \( d \) is the common difference between the terms.
This formula is incredibly useful because you can plug in the values for the first term, the term number, and the common difference to easily find any term in the sequence without having to calculate all the preceding terms.
first term
The first term of an arithmetic progression is often denoted by \( a \). It's the starting point of your sequence and plays a crucial role in determining the rest of the terms. In many arithmetic sequences, knowing the first term can help you establish a pattern and predict future terms.Consider it as your reference point. Once you know the first term and the common difference, every other term in the sequence is just a step away. This is because each subsequent term is constructed by adding the common difference to the previous term.To give an example, if your first term \( a \) is 3 and your common difference \( d \) is 2, your sequence will start with 3, and each following term will be "3 + 2", "3 + 2 + 2", and so on. In formula terms, it kicks off the entire sequence using:\[T_1 = a\]
common difference
The common difference in an arithmetic progression is the consistent difference between subsequent terms, represented by \( d \). This element is what gives the sequence its uniform structure. It tells you by how much each term increases (or decreases if \( d \) is negative) from the previous term.Understanding common difference:
  • If \( d \) is positive, each term is larger than the one before, making the sequence an increasing one.
  • If \( d \) is negative, each term is smaller than the one before, resulting in a decreasing sequence.
Calculating the common difference is straightforward. You can find \( d \) by subtracting any term in the sequence from the term that directly follows it:\[d = T_{r+1} - T_r\]Knowing the common difference is essential because, combined with the first term, it allows you to use the rth term formula to pinpoint any term in the arithmetic progression without enumerating all the previous terms.

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Most popular questions from this chapter

Let \(a, b, c, d\) and \(p\) be any non zero distinct real numbers such that \(\left(a^{2}+b^{2}+c^{2}\right) p^{2}-2(a b+b c+c d) p+\left(b^{2}+c^{2}+\right.\) \(\left.d^{2}\right)=0 .\) Then : (a) \(a, c, p\) are in A.P. (b) \(a, c, p\) are in GP. (c) \(a, b, c, d\) are in G.P. (d) \(a, b, c, d\) are in A.P.

The sum of the first 20 terms common between the series 3 \(+7+11+15+\ldots \ldots . . .\) and \(1+6+11+16+\ldots \ldots\), is |Online April 11, 2014] (a) 4000 (b) 4020 (c) 4200 (d) 4220

Let \(\alpha\) and \(\beta\) be the roots of \(x^{2}-3 x+p=0\) and \(\gamma\) and \(\delta\) be the roots of \(x^{2}-6 x+q=0\). If \(\alpha, \beta, \gamma, \delta\) form a geometric progression. Then ratio \((2 q+p):(2 q-p)\) is: \([\) Sep. \(04,2020(\mathrm{I})]\) (a) \(3: 1\) (b) \(9: 7\) (c) \(5: 3\) (d) \(33: 31\)

Let \(\mathrm{S}_{\mathrm{n}}=1+\mathrm{q}+\mathrm{q}^{2}+\ldots .+\mathrm{q}^{\mathrm{n}}\) and \(\mathrm{T}_{\mathrm{n}}=1+\left(\frac{\mathrm{q}+1}{2}\right)+\left(\frac{\mathrm{q}+1}{2}\right)^{2}+\ldots+\left(\frac{\mathrm{q}+1}{2}\right)^{\mathrm{n}}\) where \(\mathrm{q}\) is a real number and \(\mathrm{q} \neq 1\). If \({ }^{101} \mathrm{C}_{1}+{ }^{101} \mathrm{C}_{2} \mathrm{~S}_{1}+\ldots .+{ }^{101} \mathrm{C}_{101} \cdot \mathrm{S}_{100}=\alpha \mathrm{T}_{100}\), then \(\alpha\) is equal to: [Jan. 11, 2019 (II)] (a) \(2^{99}\) (b) 202 (c) 200 (d) \(2^{100}\)

Let \(a_{1}, a_{2}, a_{3} \ldots \ldots \ldots . .\) be terms on A.P. If \(\frac{a_{1}+a_{2}+\ldots \ldots \ldots a_{p}}{a_{1}+a_{2}+\ldots \ldots \ldots .+a_{q}}=\frac{p^{2}}{q^{2}}, p \neq q\), then \(\frac{a_{6}}{a_{21}}\) equals \([2006]\) (a) \(\frac{41}{11}\) (b) \(\frac{7}{2}\) (c) \(\frac{2}{7}\) (d) \(\frac{11}{41}\)
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