Question

the third term of an ap is 7 and seventh term exceeds three times the third term by 2 find the sum of first 30 terms

Answer 1

We have  a3= a1+ (k-1)d
                7= a1 +2d          ---(1)
ATQ    a7 = a1 +6d = 3*7 +2
                  a1+ 6d = 21 + 2=23  ------(2)
Subtracting  (1) from (2) we have
                                 4d= 23-7=16
                               ⇒  d=4
  substituting value in (1) we have a1= -1
  Sum of first 30 terms S30 = 30/2[-2*1+ (30-1)4]
                                        = 15*114= 1710