Question
If $\text{a}\Big(\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big),\ \Big(\frac{1}{\text{c}}+\frac{1}{\text{a}}\Big),\ \text{c}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)$ are in A.P., proved that a, b, c are in A.P.

Answer

$\text{a}\Big(\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big),\ \Big(\frac{1}{\text{c}}+\frac{1}{\text{a}}\Big),\ \text{c}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)$ are in A.P.
$\Rightarrow\text{a}\Big(\frac{1}{\text{b}}+\frac{1}{\text{c}}\Big)+1,\ \Big(\frac{1}{\text{c}}+\frac{1}{\text{a}}\Big)+1,\ \text{c}\Big(\frac{1}{\text{a}}+\frac{1}{\text{b}}\Big)+1$ are in A.P.
$\Rightarrow\Big(\frac{​​​​​​\text{ac}+​\text{ab}+​\text{bc}}{​\text{bc}}\Big),\ \Big(\frac{​\text{ab}+​\text{bc}+​\text{ac}}{​\text{ac}}\Big),\ \Big(\frac{​\text{cd}+​\text{ac}+​\text{ab}}{​\text{ab}}\Big)$ are in A.P.
$\Rightarrow\frac{1}{​\text{bc}},\ \frac{1}{​\text{ac}},\ \frac{1}{​\text{ab}}$ are in A.P.
$\Rightarrow\frac{​\text{abc}}{​\text{bc}},\ \frac{​\text{abc}}{​\text{ac}},\ \frac{​\text{abc}}{​\text{ab}}$ are in A.P.
$\Rightarrow​\text{a},​\text{b},​\text{c}$ are in A.P.

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