Determine the ap whose third term is 16 and 7 term exceeds of term by 12
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a3 = 16
=> a + 2d = 16
=> a = 16 - 2d
a7 = a5 + 12
=> a + 6d = a + 4d + 12
=> 2d= 12
=> d = 6
a = 16 - 2d
=> a = 16 - 12
=> a = 4
AP = 4, 10, 16...
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