College Algebra, Chapter 9, 9.2, Section 9.2, Problem 60

Suppose an arithmetic sequence has first term $a_1 = 1$ and fourth term $a_4 = 16$. How many terms of this sequence must be added to get 2356?

Using the formula

$a_n = a + (n - 1)d \qquad $ To solve for $d$

$\displaystyle d = \frac{a_n - a}{n - 1} = \frac{a_4 - a_1}{4 - 4} = \frac{16 - 1}{3} = \frac{15}{3} = 5$

We use this to substitute in the formula


$
\begin{equation}
\begin{aligned}

S_n =& \frac{n }{2} [2a + (n-1) d]
\\
\\
2356 =& \frac{n}{2} [2(1) + (n-1)5]
\\
\\
2356 =& \frac{n}{2} (5n -3)
\\
\\
2(2356) =& 5n^2 - 3n
\\
\\
4712 =& 5n^2-3n
\\
\\
0 =& 5n^2 -3n - 4712

\end{aligned}
\end{equation}
$


Using Quadratic Formula,

We have



$
\begin{equation}
\begin{aligned}

n =& \frac{-3 \pm \sqrt{(-3)^2 - 4(5)(-4712)}}{2(5)}
\\
\\
n =& 31 \text{ and } n = \frac{-152}{5}

\end{aligned}
\end{equation}
$


Since $n$ can't have a negative value $n = 31$

It needs $31$ terms in order to get $2356$.