if a, b, c are in A. P., prove that a(b+c) /bc, b(c+a) /ca, c(a+b) /ab are also in A. P.
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Step-by-step explanation:
a , b , c are in A.P
//divide by abc
=> a/abc , b/abc, c/abc are in A.P
=> 1/bc, 1/ac, 1/ab are in A.P.
//multiply by ab + bc+ ca
=> ab + bc + ca / bc, ab + bc + ca / ac, ab + bc + ca / ab are in AP
//subtract 1 from all
=> 1/bc(ab + bc + ca ) - 1 , 1/ac(ab + bc + ca) - 1, 1/ab(ab + bc + ca) - 1 are in A.P
=> ab+bc+ca - bc / bc , ab+bc+ca - ac / ac, ab+bc+ca - ab/ab are in A.P
=> ab + ca /bc , ab + bc/ac, bc + ca / ab are in A.P
=> a(b+c)/bc, b(a+c)/ac, c(a+b)/ab are in A.P
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