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Question
Determine the A.P. whose 3rd term is 16 and the 7th term exceeds the 5th term by 12.
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Solution
We have given a3 = 16
a + (3 − 1)d = 16
a + 2d = 16(1)
a7 − a5 = 12
[a + (7 − 1)d] − [a + (5 − 1)d] = 12
(a + 6d) − (a + 4d) = 12
2d = 12
d = 6
From equation (1), we obtain
a + 2(6) = 16
a + 12 = 16
a = 4
Therefore, A.P. will be 4, 10, 16, 22, ...
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