Divide 56 in four parts in A.P. such that the ratio of the product of their extremes [1st and 4th] to the product of mean [2nd and 3th ] is 5:6 .
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Let the no's be (a-3d),(a-d),(a+d),(a+3d)
Then,
(a-3d)+(a-d)+(a+d)+(a+3d)=56
4a=56
a=14
(a-3d)(a+3d)/(a-d)(a+d)=5/6
(a^2-9d^2)/(a^2-d^2)=5/6
Putting a=14
196-9d^2/(196-d^2)=5/6
1176-54d^2=980-5d^2
49d^2=196
d^2=4
d=+-2
The answer is 8,12,16,20
Or 20,16,12,8
Then,
(a-3d)+(a-d)+(a+d)+(a+3d)=56
4a=56
a=14
(a-3d)(a+3d)/(a-d)(a+d)=5/6
(a^2-9d^2)/(a^2-d^2)=5/6
Putting a=14
196-9d^2/(196-d^2)=5/6
1176-54d^2=980-5d^2
49d^2=196
d^2=4
d=+-2
The answer is 8,12,16,20
Or 20,16,12,8
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