If bc,ca,ab are in HP show that
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kvnmurty:
AP... a(b+c)/bc , .. etc? are bc, ac, ab in the denominators ??
reply @zubair
yes
plz give me the solution
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Use the Latex Equation Editor of Brainly...
When bc, ca, ab are in HP, it means that a, b, and c are in AP.
Use this fact to show that the three given terms are in AP.
![bc, ca\ and\ ab\ are\ in\ HP. \ So\ reciprocals\ are\ in\ AP.\\\\\frac{1}{bc}+\frac{1}{ab}=\frac{2}{ca} . =\ \textgreater \ \frac{a+c}{abc}=\frac{2b}{abc}.\ =\ \textgreater \ a+c=2b.\\a,\ b, \ c\ are\ in \ AP.\\\\Add\ first\ and\ last\ terms.\\\frac{a(b+c)}{bc}+\frac{c(a+b)}{ab}=\frac{a^2b+a^2c+c^2a+c^2b}{abc}=\frac{b(a^2+c^2)+ac(a+c)}{abc}\\\\=\frac{b[(a+c)^2-2ac]+ac(2b)}{abc} =\frac{b[4b^2-2ac]+2abc}{abc}=\frac{4b^2}{ac}\\\\2 \times Middle\ term=2 \times \frac{b(a+c)}{ac}=2 \times \frac{b*2b}{ac}=\frac{4b^2}{ac}](https://tex.z-dn.net/?f=bc%2C+ca%5C+and%5C+ab%5C+are%5C+in%5C+HP.+%5C+So%5C+reciprocals%5C+are%5C+in%5C+AP.%5C%5C%5C%5C%5Cfrac%7B1%7D%7Bbc%7D%2B%5Cfrac%7B1%7D%7Bab%7D%3D%5Cfrac%7B2%7D%7Bca%7D+.+%3D%5C+%5Ctextgreater+%5C++%5Cfrac%7Ba%2Bc%7D%7Babc%7D%3D%5Cfrac%7B2b%7D%7Babc%7D.%5C+%3D%5C+%5Ctextgreater+%5C++a%2Bc%3D2b.%5C%5Ca%2C%5C+b%2C+%5C+c%5C+are%5C+in+%5C+AP.%5C%5C%5C%5CAdd%5C+first%5C+and%5C+last%5C+terms.%5C%5C%5Cfrac%7Ba%28b%2Bc%29%7D%7Bbc%7D%2B%5Cfrac%7Bc%28a%2Bb%29%7D%7Bab%7D%3D%5Cfrac%7Ba%5E2b%2Ba%5E2c%2Bc%5E2a%2Bc%5E2b%7D%7Babc%7D%3D%5Cfrac%7Bb%28a%5E2%2Bc%5E2%29%2Bac%28a%2Bc%29%7D%7Babc%7D%5C%5C%5C%5C%3D%5Cfrac%7Bb%5B%28a%2Bc%29%5E2-2ac%5D%2Bac%282b%29%7D%7Babc%7D+%3D%5Cfrac%7Bb%5B4b%5E2-2ac%5D%2B2abc%7D%7Babc%7D%3D%5Cfrac%7B4b%5E2%7D%7Bac%7D%5C%5C%5C%5C2+%5Ctimes+Middle%5C+term%3D2+%5Ctimes+%5Cfrac%7Bb%28a%2Bc%29%7D%7Bac%7D%3D2+%5Ctimes+%5Cfrac%7Bb%2A2b%7D%7Bac%7D%3D%5Cfrac%7B4b%5E2%7D%7Bac%7D "bc, ca\ and\ ab\ are\ in\ HP. \ So\ reciprocals\ are\ in\ AP.\\\\\frac{1}{bc}+\frac{1}{ab}=\frac{2}{ca} . =\ \textgreater \ \frac{a+c}{abc}=\frac{2b}{abc}.\ =\ \textgreater \ a+c=2b.\\a,\ b, \ c\ are\ in \ AP.\\\\Add\ first\ and\ last\ terms.\\\frac{a(b+c)}{bc}+\frac{c(a+b)}{ab}=\frac{a^2b+a^2c+c^2a+c^2b}{abc}=\frac{b(a^2+c^2)+ac(a+c)}{abc}\\\\=\frac{b[(a+c)^2-2ac]+ac(2b)}{abc} =\frac{b[4b^2-2ac]+2abc}{abc}=\frac{4b^2}{ac}\\\\2 \times Middle\ term=2 \times \frac{b(a+c)}{ac}=2 \times \frac{b*2b}{ac}=\frac{4b^2}{ac}")
Hence proved.
===== another way
When bc, ca, ab are in HP, it means that a, b, and c are in AP.
Use this fact to show that the three given terms are in AP.
Hence proved.
===== another way
:-)
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