If a,b,c are in AP, prove that :
(i) $(a - c)^{2} = 4 (a - b) (b - c)$
(ii) $a^{2} + c^{2} + 4 a c = 2 (a b + b c + c a)$

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Solution

$(i) (a - c)^{2} = 4 (a - b) (b - c)$

R.H.S $4 (a - b) (b - c)$

$= 4 (a - (\frac{a + c}{2})) (\frac{a + c}{2} - c) (2 b = a + c)$

$= 4 (\frac{a - c}{2}) (\frac{a - c}{2})$

$= (a - c)^{2}$

= L.H.S

$(i i) a^{2} + c^{2} + 4 a c = 2 (a b + b c + c a)$

R.H.S $2 (a b + b c + c a)$

$= 2 (a (\frac{a + c}{2}) + (\frac{a + c}{2}) c + c a)$ $(∵$ o,b,c are in AP $∴ 2 b = a + c)$

$= 2 (\frac{a^{2} + a c + a c + c^{2} + 2 a c}{2})$

$= a^{2} + c^{2} + 4 a c$

$= L . H . S$

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