In the AP (31,45,59,\ldots), what is the greatest term between (900) and (1000)?

Correct answer: \(997\)

Here, the first term is \(a=31\) and the common difference is \(d=14\). Thus, \(a_n=31+14(n-1)\). To find the greatest term below \(1000\), use \(31+14(n-1)<1000\), which gives \(n-1\leq69\). Therefore, the term is \(31+14\times69=997\). The next term is \(1011\), which exceeds \(1000\). Exam tip: for the greatest AP term below a limit, use an inequality and take the greatest possible integer value of \(n-1\).