Let the A.P. be a, a + d, a + 2d, a + 3d, ... a + (n – 2) d, a + (n – 1)d.
Sum of first four terms = a + (a + d) + (a + 2d) + (a + 3d) = 4a + 6d
Sum of last four terms = [a + (n – 4) d] + [a + (n – 3) d] + [a + (n – 2) d] + [a + n – 1) d] = 4a + (4n – 10) d
According to the given condition,
4a + 6d = 56
⇒ 4(11) + 6d = 56 [Since a = 11 (given)]
⇒ 6d = 12
⇒ d = 2
∴ 4a + (4n –10) d = 112
⇒ 4(11) + (4n – 10)2 = 112
⇒ (4n – 10)2 = 68
⇒ 4n – 10 = 34
⇒ 4n = 44
⇒ n = 11
Thus, the number of terms of the A.P. is 11.