Question

if a b c are in ap then show that a^2(b+c) b^2(c+a) c^2(a+b) are also in ap

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Answer 1

$\bold{∵a,b,c \: are \: in AP}$

$\bold{hence,(b−a)=(c−b)⟶(1)}$

$\bold{(i) ∵b^2 (c+a)−a^2 (b+c)}$

$\bold{=(b−a)(ab+bc+ca)}$

$\bold{and \: c^2 (a+b)−b^2 (c+a)}$

$\bold{=(c−b)(ab+bc+ca)}$

$\bold{=(b−a)(ab+bc+ca)(from(1)}$

${ \boxed{ \small \bold{hence,b^2 (c+a)−a^2 (b+c)=c^2 (a+b)−b^2 (c+a)}}}$

$\small \bold{∴a^2(b+c),b^2 (c+a),c^2(a+b)are \: also \: in \: AP}$