If frac{1}{b - a} + frac{1}{b - c} = frac{1}{ | vedclass

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(C) Given the equation: $\frac{1}{b - a} + \frac{1}{b - c} = \frac{1}{a} + \frac{1}{c}$.Simplify the left side: $\frac{(b - c) + (b - a)}{(b - a)(b -

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Harmonic Progression

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(C) Given the equation: $\frac{1}{b - a} + \frac{1}{b - c} = \frac{1}{a} + \frac{1}{c}$.Simplify the left side: $\frac{(b - c) + (b - a)}{(b - a)(b - c)} = \frac{2b - a - c}{b^2 - ab - bc + ac}$.Simplify the right side: $\frac{a + c}{ac}$.Equating both sides: $\frac{2b - a - c}{b^2 - b(a + c) + ac} = \frac{a + c}{ac}$.Cross-multiplying: $ac(2b - a - c) = (a + c)(b^2 - b(a + c) + ac)$.$2abc - a^2c - ac^2 = ab^2 - ab(a + c) + a^2c + cb^2 - cb(a + c) + ac^2$.$2abc - a^2c - ac^2 = b^2(a + c) - b(a + c)^2 + a^2c + ac^2$.$2abc = b^2(a + c) - b(a + c)^2 + 2a^2c + 2ac^2$.$2abc = b^2(a + c) - b(a + c)^2 + 2ac(a + c)$.Dividing by $(a + c)$ (assuming $a + c 
eq 0$): $\frac{2abc}{a + c} = b^2 - b(a + c) + 2ac$.$2abc = b^2(a + c) - b(a + c)^2 + 2ac(a + c)$.This simplifies to $b = \frac{2ac}{a + c}$,which is the condition for $a, b, c$ to be in Harmonic Progression.

If $\frac{1}{b - a} + \frac{1}{b - c} = \frac{1}{a} + \frac{1}{c}$,then what is the relationship between $a, b, c$?

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