If $a,b,c$ are in harmonic progression, then the value of $\frac{a-b}{b-c}$ is equal to?

Solution

1. Since $a,b,c$ are in harmonic progression, their reciprocals $\frac{1}{a},\frac{1}{b},\frac{1}{c}$ form an arithmetic progression, implying $\frac{2}{b}=\frac{1}{a}+\frac{1}{c}$.
2. Solving for $b$, we get $\frac{2}{b}=\frac{a+c}{ac}$, which simplifies to $b=\frac{2ac}{a+c}$.
3. Substitute $b$ into the expression $\frac{a-b}{b-c}$ to get $\frac{a-\frac{2ac}{a+c}}{\frac{2ac}{a+c}-c}$.
4. Simplify the numerator to $\frac{a^2+ac-2ac}{a+c}=\frac{a(a-c)}{a+c}$ and the denominator to $\frac{2ac-ac-c^2}{a+c}=\frac{c(a-c)}{a+c}$.
5. Dividing the two expressions yields $\frac{a(a-c)}{c(a-c)}=\frac{a}{c}$.
6. Hence the answer is (A).