Finding an Arithmetic Sequence Given a Relation Between 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
\[- \frac{5}{2}, \: -1, \: \frac{1}{2}, \: 2, \: \frac{7}{2}, \: 5\]
