a, b and c are in A.P. Show that: $(b + c - a), (c + a - b)$ and $(a + b - c)$ are also in A.P.

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Solution

$a, b, c$ are in A.P
$2 b = a + c 2 (c + a - b) = 2 (2 b - b) = 2 b = (a + 2 b + c - a - c) 2 (c + a - b) = (b + c - a) + (a + b - c)$
$(b + c - a), (c + a - b), (a + b - c)$ are in A.P

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