If \(a_n=\frac{n}{2n-1}\), which is the first term less than \(\frac{3}{5}\)?

Correct answer: Fourth term

Solve \(\frac{n}{2n-1}<\frac{3}{5}\). For \(n\geq1\), \(2n-1\) is positive, so cross-multiplication gives \(5n<3(2n-1)\), or \(5n<6n-3\). Hence, \(n>3\). The smallest integer value of \(n\) is 4, so the first such term is \(a_4\), the fourth term. Note that \(a_3=\frac{3}{5}\), which is equal to, not less than, \(\frac{3}{5}\). Exam tip: For “less than”, use \(<\); an equal value is not included.