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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
For given A.P
`t_3 = a + 2d = 16` ....(i)
Now
`t_7 - t_5 = 12`
`=> (a + 6d) - (a + 4d) = 12`
⇒ 2d = 12
⇒ d = 6
Substituting the value of din (i) we get
`a + 2 xx 6 = 16`
⇒ a + 12 = 16
`=> a = 4`
Thus the required A.P = a, a + d, a + 2d, a + 3d,..
= 4, 10, 16, 22,...
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