determine the AP whose 3rd terms 16and 7th term exceeds the 5th term by12
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3rd term 16
7th term = 5th+ 12
a +(3-1)d= a+2d = 16 1st equ.
a+6d = a+4d+ 12 2nd equ.
from 2nd
2d = 12
d = 6
putting the value of d in 1st equ.
a+ 2×6 = 16
a = 4
therefore, ap is 4,10,16,22,28,34......
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