Question 936199
formula:
  nth term  =a+(n-1)d
 sum of n terms = {{{(n/2)(2a+(n-1)d)}}}
 where d = common difference , a = first term
 9 th term  =  a+(9-1)d
    52 = a+8d...................eq(1)
  sum of fisrt 12 terms = 414
   {{{(12/2)(2a+(12-1)d) =414}}}
  {{{6(2a+11d)=414}}}
     divide with 6 on both sides
 {{{ 6(2a+11d)/6 =414/6}}}
 {{{2a+11d = 69}}}.............eq(2)
 we need to solve eq(1) & eq(2) 
 *[invoke linear_substitution "a", "d", 1, 8, 52, 2, 11, 69 ]