## Finding an Arithmetic Sequence Given a Relation Between Terms

A certain arithmetic sequence has 3rd, 4th and 6th terms
$\frac{1}{k}, \: k, \: k^2+1$

$k \in \mathbb{Q}$
.
Because the sequence is arithmetic, the difference between any two successive terms, called the common difference
$d$
, is the same. Hence
$d= k- \frac{1}{k}, \: 2d=k^2+1-k \rightarrow 2k-\frac{2}{k}=k^2+1-k \rightarrow k^3-3k^2+k+2=0$
. This cubic expression factorises and we get
$(k-2)(k^2-k-1)=0$
.
Since
$k \in \mathbb{Q}$
,
$k=2$
and the common difference is
$d=k-\frac{1}{k}=2- \frac{1}{2}= \frac{3}{2}$
.
The first six terms are
$- \frac{5}{2}, \: -1, \: \frac{1}{2}, \: 2, \: \frac{7}{2}, \: 5$