91Ó°ÊÓ

Suppose that there is a linear function \(f(n) = mn + b\), where \(m\) and \(b\) are constants, such that \(a_{n} = f(n)\) for every positive integer \(n\). We need to show that this sequence is an arithmetic sequence. By the definition of an arithmetic sequence, the difference between successive terms should be constant. Let's test this by finding \(a_{n+1} - a_{n}\): \[ a_{n+1} - a_{n} = f(n+1) - f(n) = m(n+1) + b - (mn + b) = mn + m + b - mn - b = m, \] which is constant as \(n\) varies. Therefore, if a sequence is represented by a linear function, it's an arithmetic sequence.

Suppose the infinite sequence \(a_{1}, a_{2}, a_{3}, \ldots\) is an arithmetic sequence. We want to find a linear function \(f(n) = mn + b\) that generates this sequence. Since the sequence is an arithmetic sequence, there is a constant difference between the terms, say \(d\). That is, \(a_{n+1} - a_{n} = d\). The first term in this sequence is \(a_{1}\), so let's define our desired linear function to be \(f(n) = d(n-1) + a_{1}\), for \(n \ge 1\). Let's check if this function generates the sequence. We have: \[ f(1) = d(1-1) + a_{1} = a_{1} \quad\text{-- this is our first term.} \] So far so good, let's continue: \[ f(2) = d(2-1) + a_{1} = d + a_{1} = a_{1} + d = a_{2} \quad\text{-- this is our second term.} \] And we see that every \(f(n)\) adds another \(d\) to \(a_{1}\), which by definition of an arithmetic sequence gives us exactly the terms of the sequence. In other words, if a sequence is an arithmetic sequence, then it can be represented by a linear function. By providing the proofs for both directions, we have shown that an infinite sequence \(a_{1}, a_{2}, a_{3}, \ldots\) is an arithmetic sequence if and only if there is a linear function \(f\) such that \(a_{n}=f(n)\) for every positive integer \(n\).