If (2a+3, 5a-1, 11a-13) are in an arithmetic progression, what is the second term?

Correct answer: \(\frac{37}{3}\)

In an arithmetic progression, twice the middle term equals the sum of the first and third terms. Thus, \(2(5a-1)=(2a+3)+(11a-13)\). This gives \(10a-2=13a-10\), so \(a=\frac{8}{3}\). Hence, the second term is \(5a-1=5\times\frac{8}{3}-1=\frac{37}{3}\). The value \(14\) results from the incorrect calculation \(a=3\). Exam tip: for three AP terms, use \(2b=a+c\) directly.