In the AP (156,142,128,\ldots), which is the last term greater than (-120)?

Correct answer: (-110)

Here, the first term is \(a=156\) and the common difference is \(d=-14\). The \(n\)th term is \(a_n=156+(n-1)(-14)=170-14n\). For a term greater than \((-120)\), \(170-14n>-120\), so \(n<20.71\). Thus the greatest integral value is \(n=20\), giving \(a_{20}=-110\). The next term, \(a_{21}=-124\), is not greater than \((-120)\). Exam tip: In a decreasing AP, always check the next term after finding the candidate last term.