Asked by Vijay
Divide 24 into four parts which are in A.P. such that the sum of their squares is 164.
Answers
Answered by
Reiny
let the four parts be
a, a+d, a+2d, and a+3d
a + a+d + a+2d + a+3d= 24
4a + 6d = 24
2a + 3d = 12 ----> d = (12-2a)/3
a^2 + (a+d)^2 + (a+2d)^2 + (a+3d)^2 = 164
4a^2 + 12ad + 14d^2 = 164
2a^2 + 6ad + 7d^2 = 82
2a^2 + 6a(12-2a)/3 + 7(12-2a)^2/9 = 82
times 9
18a^2 + 216a - 36a^2 + 1008 - 336a + 28a^2 = 738
10a^2 -120a + 270 = 0
a^2 - 12a + 27 = 0
(a-9)(a-3) = 0
a = 9 or a = 3
if a=9, d = -2, AP is 9, 7, 5, and 3
if a=3, d = 2 , the AP is 3,5,7, and 9
(notice the numbers are just reversed)
a, a+d, a+2d, and a+3d
a + a+d + a+2d + a+3d= 24
4a + 6d = 24
2a + 3d = 12 ----> d = (12-2a)/3
a^2 + (a+d)^2 + (a+2d)^2 + (a+3d)^2 = 164
4a^2 + 12ad + 14d^2 = 164
2a^2 + 6ad + 7d^2 = 82
2a^2 + 6a(12-2a)/3 + 7(12-2a)^2/9 = 82
times 9
18a^2 + 216a - 36a^2 + 1008 - 336a + 28a^2 = 738
10a^2 -120a + 270 = 0
a^2 - 12a + 27 = 0
(a-9)(a-3) = 0
a = 9 or a = 3
if a=9, d = -2, AP is 9, 7, 5, and 3
if a=3, d = 2 , the AP is 3,5,7, and 9
(notice the numbers are just reversed)
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