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If $\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{a}+\frac{1}{c}$, then $a, b, c$ are in
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Verified Answer
The correct answer is:
$\mathrm{HP}$
Suppose a, $\mathrm{b}$ and $\mathrm{c}$ are in $\mathrm{HP}$.
$b=\frac{2 a c}{a+c}$
Now, consider
$\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{\frac{2 a c}{a+c}-a}+\frac{1}{\frac{2 a c}{a+c}-c}$
$=\frac{a+c}{a(c-a)}+\frac{a+c}{c(a-c)}=\left(\frac{a+c}{c-a}\right)\left(\frac{1}{a}-\frac{1}{c}\right)$
$=\frac{a+c}{c-a} \times \frac{c-a}{c a}=\frac{a+c}{c a}=\frac{1}{a}+\frac{1}{c}$
Thus, our supposition is correct. Hence, $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ are in $\mathrm{HP}$.
$b=\frac{2 a c}{a+c}$
Now, consider
$\frac{1}{b-a}+\frac{1}{b-c}=\frac{1}{\frac{2 a c}{a+c}-a}+\frac{1}{\frac{2 a c}{a+c}-c}$
$=\frac{a+c}{a(c-a)}+\frac{a+c}{c(a-c)}=\left(\frac{a+c}{c-a}\right)\left(\frac{1}{a}-\frac{1}{c}\right)$
$=\frac{a+c}{c-a} \times \frac{c-a}{c a}=\frac{a+c}{c a}=\frac{1}{a}+\frac{1}{c}$
Thus, our supposition is correct. Hence, $\mathrm{a}, \mathrm{b}$ and $\mathrm{c}$ are in $\mathrm{HP}$.
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