Given that four positive numbers a, B, C and D satisfy a + D = B + C, b square = AC, and two thirds of C equals one part of B + one part of D, try to find a: B: C: D It's a detailed process

Given that four positive numbers a, B, C and D satisfy a + D = B + C, b square = AC, and two thirds of C equals one part of B + one part of D, try to find a: B: C: D It's a detailed process

solution
Because B & # 178; = AC
So a, B, C are in the same ratio sequence,
Let the common ratio be K, then B = AK, C = AK & # 178;
From a + D = B + C
d=b+c-a=（k²+k-1）a
From "two of C equals one of B + one of D"
2/c=1/b+1/d=(b+d)/bd
2bd=c(b+d)
2k（k²+k-1）a²=k²[k+（k²+k-1）]
Simplify and divide a by the same number to get (a is a positive number, so it can be divided by the same number)
k^3-3k+2=0
The solution is k = 1 or K = - 2
Because a, B, C and D are all positive numbers,
Therefore, k = - 2 does not conform to the meaning of the question and should be discarded
So k = 1
So B = C = a
d=（k²+k-1）a=a
So a = b = C = D
So a: B: C: D = 1:1:1:1