Find the arithmetic progression whose third term is 16 and seventh term exceeds its fifth term by 12.

Given, a3 = 16

a + 2d = 16 (i)


a5 = a + (5 – 1) d


= a + 4d


a7 = a + (7 -1)d


= a + 6d


Acc. To question,


a7 = 12 + a5 (ii)


Put the value of a5 and a7 in (ii), we get


a + 6d = 12 + a + 4d


2d = 12 + a – a


2d = 12


d = 6


From equation (i),


16 = a + 2(6)


16 = a + 12


a = 4


We have to find A.P.


a1 = 4


a2 = a + d = 4 + 6 = 10


a3 = a + 2d = 4 + 2(6) = 16


a4 = a + 3d = 4 + 3(6) = 22


Hence, the A.P. is 4 , 10, 16 , 22,…


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