Determine the AP whose third term is 16 and the 7thterm exceeds the 5thterm by 12.
Given
a3 = 16; a7 = a5 + 12
a + (7-1) d = a + (5-1) d +12
2d = 12
d = 12
Common difference (d) = 12
Substituting value of d in a3
a3 = 16
a + (3-1) d = 16
a + 2 (6) = 16
a = 4
∴ The series will be a, a+d, a+2d, a+3d,….
4, 10, 16, 22,…