Question

The first and the last terms of an AP are 7 and 49respectively. If sum of all its terms is 420, find its common difference

Answer 1

Given,
                              a = 7 ---------------- (i)
             a + (n - 1)d = 49----------------(ii)
 n/2[a + a +(n - 1)d] = 420--------------(iii)

  By (i) and (ii) we get,
              7 + (n - 1)d = 49  ----------------(iv)

Similarly,
 By (i) and (iii) we have,
    n/2[7 + 7 + (n - 1)d] = 420
                   n/2[7 + 49] = 420           (From iv)
                         n/2(56) = 420
                               28n = 420 
                                   n = 420/28
                                      = 15---------------(v)

Substituting a = 7 & n = 15 [From (i) & (v)] in (ii)
   a + (n - 1)d = 49
         7 + 14d = 49
                14d = 42
                    d = 42/14
                       = 3