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Question
If a, b, c is in A.P., prove that:
a2 + c2 + 4ac = 2 (ab + bc + ca)
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Solution
Since a, b, c are in A.P., we have:
2b = a+c
\[\Rightarrow\] b = \[\frac{a + c}{2}\]
Consider RHS:
2 (ab + bc + ca)
\[\text { Substituting b } = \frac{a + c}{2}: \]
\[ \Rightarrow 2\left\{ a\left( \frac{a + c}{2} \right) + c\left( \frac{a + c}{2} \right) + ac \right\}\]
\[ \Rightarrow 2\left\{ \frac{a^2 + ac + ac + c^2 + 2ac}{2} \right\}\]
\[ \Rightarrow a^2 + 4ac + c^2 \]
\[\text { Hence, proved } .\]
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