Question 736363
For {{{arithmetic}}} sequences, the common difference is {{{d}}}, and the first term {{{a[1]}}} is often referred to simply as "{{{a}}}". Since you get the next term by adding the common difference, the value of {{{a[2]}}} is just {{{a + d}}}. The third term is {{{a[3] = (a + d) + d = a + 2d}}}. The fourth term is {{{a[4] = (a + 2d) + d = a + 3d}}}. Following this pattern, the {{{n-th}}} term an will have the form 

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

For {{{geometric}}} sequences, the common ratio is {{{r}}}, and the first term {{{a[1]}}} is often referred to simply as "{{{a}}}". Since you get the next term by multiplying by the common ratio, the value of {{{a[2]}}} is just {{{ar}}}. The third term is {{{a[3] = r(ar) = ar^2}}}. The fourth term is {{{a[4] = r(ar^2) = ar^3}}}. Following this pattern, the {{{n-th}}} term an will have the form 

{{{a[n] = ar^(n -1)}}}.

in your case  {{{12}}},{{{40}}},{{{68}}},{{{96}}} sequence is have the form 

common difference is {{{28}}} and the greatest common divisor is {{{4}}}

first term is {{{12}}} and if divided by {{{4}}}, we get {{{3}}} that could be written as {{{7n-4}}} starting with {{{n=1}}} and up
 
so, the recursive formula is:

{{{a[n]=4(7n-4)}}}

and this is {{{arithmetic}}} sequence