In the AP (240,222,204,\ldots), which is the last term greater than (-150)?

Correct answer: (-138)

Here, the first term is \(a=240\) and the common difference is \(d=-18\). Thus, \(a_n=240-18(n-1)\). Using \(a_n>-150\), we get \(240-18(n-1)>-150\), so \(n-1<\frac{65}{3}\). Therefore, the greatest possible integer is \(n=22\), and \(a_{22}=240-18(21)=-138\). The next term, \(a_{23}=-156\), is not greater than \(-150\). Exam tip: For a decreasing AP, solve the inequality and take the greatest valid integer value of \(n\).