If a, b, c are in A.P., prove that: a^2+c^2+4ac=2(ab+bc+ca) - Vidyadip

Question

If a, b, c are in A.P., prove that:
$\text{a}^2+\text{c}^2+4\text{ac}=2(\text{ab}+\text{bc}+\text{ca})$

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Answer

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

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