Which is the general term of the arithmetic progression (5,9,13,17,\ldots)?

Correct answer: \(a_n=4n+1\)

Here, the first term is \(a=5\) and the common difference is \(d=9-5=4\). Thus, \(a_n=a+(n-1)d=5+4(n-1)=4n+1\). In \(a_n=4n-1\), putting \(n=1\) gives 3 as the first term, so it is not correct. Exam tip: always substitute \(n=1\) in a general term to verify the first term.