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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  
\[- \frac{5}{2}, \: -1, \: \frac{1}{2}, \: 2, \: \frac{7}{2}, \: 5\]

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