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,…


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