It is known that 1 of a, 1 of B and 1 of C are equal difference sequence. It is proved that B + C of a, a + C of B and a + B of C are equal difference sequence

Because: (B + C) / a = (a + B + C) / A-1
(a+c)/b=(a+b+c)/b-1
(a+b)/c=(a+b+c)/c-1
So: 2 (a + C) / b = 2 (a + B + C) / B-2
=(a+b+c)×2/b-2
=(a+b+c)×(1/a+1/c)-2
=(b+c)/a+(a+b)/c

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