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