Question

if a^2 + 2bc, b^2 + 2ac,c^2+2ab are in A.P show that 1/(b-c),1/(c-a),1/(a-b) are in A.P​

Answer 1

$\sf \: 1/ b-c , 1/ c-a , 1/ a-b \\ \sf \: are \: in \: AP \\ \sf \: 2/ c-a = 1/a-b + 1/ b-c \\ \sf \: \sf \: = b-c + a-b / (a-b)(b-c)\\ \sf \: 2/ c-a = (a-c )/ (a-b )(b-c) \\ \sf \: 2(a - b)(b - c) = (c - a) {}^{2} \\ \sf \: 2ab - 2ac - 2 {b}^{2} + 2bc = { - c}^{2} { - a}^{2} + 2ac \\ \sf \: 2( {b}^{2} + 2ac) = ( {c}^{2} + 2ab + {a}^{2} + 2bc) \\ \sf \: which \: means  \\ \sf \: {a}^{2} - 2bc, {b}^{2} + 2ac , {c}^{2} + 2ab\\ \sf \: which \: is \: true$

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