Question
If a, b, c are in A.P., prove that: $(\text{a}-\text{c})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$

Answer

If $(\text{a}-\text{c})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$ Then, $\text{a}^2+\text{c}^2-2\text{ac}=4(\text{ab})-\text{b}^2-\text{ac}+\text{bc}$ $\Rightarrow\text{a}^2+\text{c}^24\text{b}^2+2\text{ac}-4\text{ac}-4\text{bc}=0$ $\Rightarrow(\text{a}+\text{c}-2\text{b})^2=0$ $\big[$ Using $(\text{a}+\text{b}+\text{c})^2=\text{a}^2+\text{b}^2+\text{c}^2+2\text{ab}+2\text{ac}+2\text{bc}\big]$ $\therefore\text{a}+\text{c}-2\text{b}=0$ or $\text{a}+\text{c}=2\text{b}$ and since, $\text{a},\ \text{b},\ \text{c}$ are in A.P [Given] $\text{a}+\text{c}=2\text{b}$ Hence proved $(\text{a}-\text{b})^2=4(\text{a}-\text{b})(\text{b}-\text{c})$

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