Question 196139
<font size = 8 color = "red"><b>Edwin's solution:</font></b>

{{{a[1]=-2&1/2}}}, {{{a[2]=-2}}}, {{{a[3]=-1&1/2}}}, {{{a[4]=-1}}}, {{{a[5]=1/2}}},...
<pre><font size = 4 color="indigo"><b>
It is an arithmetic sequence because obviously {{{1/2}}} is 
being added to each term to get the next term.  Therefore
the common difference is {{{d=1/2}}}.

The {{{n}}}th term of an arithmetic sequence is given by the
formula:

{{{a[n]=a[1]+(n-1)d}}}

So, upon substituting:

{{{a[n]=(-2&1/2)+(n-1)(1/2)}}}

Change the mixed fraction {{{-2&1/2}}} 
to the improper fraction {{{-5/2}}}

{{{a[n]=-5/2+(n-1)(1/2)}}}

Multiply through by 2 to clear of fractions:

{{{2a[n]=-5+(n-1)*1}}}

{{{2a[n]=-5+n-1}}}

{{{2a[n]=n-6}}}

Divide both sides by 2

{{{a[n]=(n-6)/2}}} 

Edwin</pre>