The fifth term of an arithmetic sequence is 9 and the 32nd term is \(60\). What is the 23rd term?

Guest May 3, 2022

#1**+2 **

Let \(a_n\) denote the \(n\)th term of this arithmetic sequence. There are 27 terms occurring from \(a_5 = 9\) to \(a_{32} = 60\). Assuming a constant common difference between terms, then the terms increase by \(\frac{60-9}{27} = \frac{17}{9}\) with each \(n\) step. That is, \(a_{n+1} = a_n + \frac{17}{9}\).

There are 18 terms occurring from \(a_5 = 9\) to \(a_{23}.\) Then, there would be 18 additions of \(\frac{17}{9}\) from \(a_5\) to \(a_{23}\). That is,

\(a_{23} = a_5 + 18\left(\frac{17}{9}\right) = 9 + 34 = 43\).

Anthrax May 3, 2022