Divide 45 in to five parts which are in a.p such that the product of extremes is 65.
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Let the four parts be (a – 3d), (a – d), (a + d) and (a + 3d).
Then, sum = 56
⇒ (a – 3d) + (a – d) + (a + d) + (a + 3d) = 56
⇒ 4a = 56
⇒ a = 14
It is given that
Thus, the four parts are a – 3d, a – d, a + d, a + 3d i.e., 8, 12, 16 and 20.
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