College Algebra, Chapter 9, 9.2, Section 9.2, Problem 48

Determine the partial sum $s_u$ of the arithmetic sequence that satisfies $a_2 = 8, a_5 = 9.5$ and $n = 15$.

First, we need to find the common difference of the sequence to find the first term. So,
For $a_2$,

$\begin{equation} \begin{aligned} a_n &= a + (n - 1) d\\ \\ a_2 &= a + (2 - 1) d\\ \\ 8 &= a + d && \text{Equation 1} \end{aligned} \end{equation}$

For $a_5$,

$\begin{equation} \begin{aligned} a_5 &= a + (5 -1 ) d\\ \\ 9.5 &= a + 4d && \text{Equation 2} \end{aligned} \end{equation}$

Using Equations 1 and 2, we get

$\begin{equation} \begin{aligned} 8 -d &= 9.5 - 4d \\ \\ 3d &= 1.5\\ \\ d &= 0.5 \end{aligned} \end{equation}$

Thus, the first term is

$\begin{equation} \begin{aligned} 8 &= a + 0.5\\ \\ a &= 7.5 \end{aligned} \end{equation}$

Therefore, the sum is

$\begin{equation} \begin{aligned} s_n &= \frac{n}{2}[2a + (n - 1) d]\\ \\ s_{15} &= \frac{15}{2} [ 2 (7.5) + (15 -1) (0.5)]\\ \\ &= 165 \end{aligned} \end{equation}$