Determine the AP whose third term is 16 and the 7th term exceeds the 5th term by 12.


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Solution: As the 7th term exceeds the 5th term by 12, so the 5th term will exceed the 3rd term by 12 as well
So, n3 = 16
n5 = 28
n7 = 40
n4 or n6 can be calculated by taking an average of the preceding and next term
So, n4 = (28+16)/2 = 22

This gives the d = 6
AP: 4, 10, 16, 22, 28, 34, 40, 46, ……..

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