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. b W& 4 a “j}boo in AW s it |
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Answer» a(1/b+1/c) + c(1/a+1/b)=a(c +b)/bc + c(a+b)/ab={a²(c +b) + c²(a+b)}/abc={a²c +a²b + c²a+c²b)}/abc={a²c+ c²a +a²b +c²b)}/abc={ac(a+c) +a²b +c²b)}/abc={ac(2b) +a²b +c²b)}/abc Since a,b,c are in AP hence a+c=2b={2abc +a²b +c²b)}/abc=b{2ac +a² +c²)}/abc=b{(a+c)²)}/abc={(a+c)²)}/ac={(a+c)(a+c))}/ac=(2b)(a+c))}/ac Since a,b,c are in AP hence a+c=2b=(2b)(1/c+1/a)hence a(1/b+1/c), b(1/c+1/a), c(1/a+1/b) are in A.P if a,b,c are in AP |
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