determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12
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a 3 = 16 = a + 2d = 16 -------------- 1
a 7 - a 5 =12
a +6 d - ( a + 4 d ) = 12
a + 6d -a - 4d = 12
6d - 4d = 12
2d = 12
d = 12 / 2
d = 6
substituting d = 6 in equation 1
a+2d = 16
a + 2 ( 2) = 16
a + 4 = 16
a = 16 - 4
a = 12
The AP so formed is
12 ,18 ,24 ,30
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