Question 975031
{{{1-3/4=1/4}}} and {{{3/4-1/2=1/4}}} ,
so the common difference of this A.P. (Arithmetic Progression, or arithmetic sequence) is
{{{d=1/4}}} .
The first term is, obviously {{{a[1]=1/2}}} .
So, as for all AP's the nth term is
{{{a[n]=a[1]+(n-1)*d}}} , 
in this case
{{{a[n]=1/2+(n-1)(1/4)}}} .
It may be easier to calculate as
{{{a[n]=(n+1)(1/4)}}} , because
{{{a[n]=1/2+(n-1)(1/4)=2/4+(n-1)(1/4)=2(1/4)+(n-1)(1/4)=(2+n-1)(1/4)=(n+1)(1/4)}}}