Question 141070
Look here for some hints http://en.wikipedia.org/wiki/Arithmetic_progression

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

Let the number of terms be {{{2k+1}}}

{{{0 = n( 2a[1] + (2k+1-1)d )/ 2 }}} 
{{{0 = n( 2a[1] + (2k)d )/ 2 }}} 
{{{0 = n( a[1] + (k)d ) }}} 
-kd = a[1]

Now the function for the n[th] term of series is
{{{a[n] = a[1] + (n - 1)d}}}
{{{a[n] = -kd + (n - 1)d}}}
{{{a[n] = (n - 1 -k)d}}}
{{{a[k+1]   = (k +1 - 1 -k)d}}}
{{{a[k+1]   = (0)d}}}
{{{a[k+1] = 0}}}