Find four numbers in AP such that their sum is 24 and product is 945.
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Let the terms be (a - 3d), (a - d), (a + d), (a + 3d)
Given that sum is 24
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 24
= 4a = 24
= a = 6
Product is 945 with a = 6
(6 - 3d)(6 - d)(6 + d)(6 + 3d) = 945
(36 - 9d^2)(36 - d^2) = 945
OR
9d^4 - 360d^2 + 351 = 0
= d = 1, 6.24
Taking the integer part (d = 1) and a = 6 gives the AP as
(6 - 3), (6 - 1), (6 + 1), (6 + 3)
= 3, 5, 7, 9
Given that sum is 24
(a - 3d) + (a - d) + (a + d) + (a + 3d) = 24
= 4a = 24
= a = 6
Product is 945 with a = 6
(6 - 3d)(6 - d)(6 + d)(6 + 3d) = 945
(36 - 9d^2)(36 - d^2) = 945
OR
9d^4 - 360d^2 + 351 = 0
= d = 1, 6.24
Taking the integer part (d = 1) and a = 6 gives the AP as
(6 - 3), (6 - 1), (6 + 1), (6 + 3)
= 3, 5, 7, 9
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