Wednesday, March 22, 2017

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}
$

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