nth term of an arithmetic sequence is given by Tn = a + (n - 1)d
6th term => a + (6 - 1)d = 8 => a + 5d = 8 . . . (1)
10th term => a + (10 - 1) d = 13 => a + 9d = 13 . . . (2)
(2) - (1) => 4d = 5 => d = 5/4
From (1), a + 5(5/4) = 8 => a + 25/4 = 8 => a = 8 - 25/4 = 7/4
Sum of n terms of an A.P. is given by Sn = n/2 [2a + (n - 1)d]
S4 = 4/2 [2(7/4) + (4 - 1)5/4] = 2[7/2 + 3(5/4)] = 2[7/2 + 15/4] = 2(29/4) =14.5
Therefore, sum of the first 4 terms is 14.5
