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Arithmetic Progressions — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsArithmetic Progressions5 Marks
Question
If a, b, c are in A.P., prove that:
$\text{a}^2+\text{c}^2+4\text{ac}=2(\text{ab}+\text{bc}+\text{ca})$
✓

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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