Determine the A.P. whose third term is 16 and the 7th term exceeds the 5th term by 12.
a3 = 16
a + (3 − 1) d = 16
a + 2d = 16 ... (i)
a7 − a5 = 12
[a+ (7 − 1) d] − [a + (5 − 1) d]= 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (i), we get,
a + 2 (6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be
4, 10, 16, 22, …
