# The fifth term of an arithmetic series is 9, and the sum of the first 16 terms is 480. How do you find the first three terms of the sequence?

Jan 24, 2016

Solve to find the first term and common difference of the sequence and hence the first three terms:

$\frac{15}{7}$, $\frac{27}{7}$, $\frac{39}{7}$

#### Explanation:

The general term of an arithmetic sequence is given by the formula:

${a}_{n} = a + d \left(n - 1\right)$

where $a$ is the initial term and $d$ is the common difference.

The sum of $N$ consecutive terms of an arithmetic sequence is $N$ times the average of the first and last terms, so:

${\sum}_{n = 1}^{16} {a}_{n} = 16 \frac{{a}_{1} + {a}_{16}}{2} = 8 \left(a + \left(a + 15 d\right)\right) = 16 a + 120 d$

From the conditions of the question we have:

$240 = 16 a + 120 d$

$9 = {a}_{5} = a + 4 d$

Hence:

$96 = 240 - 16 \cdot 9 = \left(16 a + 120 d\right) - 16 \left(a + 4 d\right)$

$= \textcolor{red}{\cancel{\textcolor{b l a c k}{16 a}}} + 120 d - \textcolor{red}{\cancel{\textcolor{b l a c k}{16 a}}} - 64 d = 56 d$

So $d = \frac{96}{56} = \frac{12}{7}$

and $a = 9 - 4 d = 9 - \frac{48}{7} = \frac{15}{7}$

Hence the first three terms of the series are:

$\frac{15}{7}$, $\frac{27}{7}$, $\frac{39}{7}$