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If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between

    \[({{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}})\text{ }\mathbf{and}\text{ }({{\mathbf{b}}^{\mathbf{2}}}~+\text{ }{{\mathbf{c}}^{\mathbf{2}}})\]

Class 10 | Exercise 1 | ICSE | Maths | ML Aggarwal | Ratio and Proportion
If b is the mean proportional between a and c, prove that (ab + bc) is the mean proportional between

    \[({{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}})\text{ }\mathbf{and}\text{ }({{\mathbf{b}}^{\mathbf{2}}}~+\text{ }{{\mathbf{c}}^{\mathbf{2}}})\]

It is given that

b is the mean proportional between a and c

    \[{{b}^{2}}~=\text{ }ac\]

…. (1)

Here (ab + bc) is the mean proportional between

    \[({{\mathbf{a}}^{\mathbf{2}}}~+\text{ }{{\mathbf{b}}^{\mathbf{2}}})\text{ }\mathbf{and}\text{ }({{\mathbf{b}}^{\mathbf{2}}}~+\text{ }{{\mathbf{c}}^{\mathbf{2}}})\]

    \[{{\left( ab\text{ }+\text{ }bc \right)}^{2}}~=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}})\text{ }({{b}^{2}}~+\text{ }{{c}^{2}})\]

Consider LHS =

    \[{{\left( ab\text{ }+\text{ }bc \right)}^{2}}\]

Expanding using formula

    \[=\text{ }{{a}^{2}}{{b}^{2}}~+\text{ }{{b}^{2}}{{c}^{2}}~+\text{ }2a{{b}^{2}}c\]

Using equation (1)

    \[\begin{array}{*{35}{l}} =\text{ }{{a}^{2}}~\left( ac \right)\text{ }+\text{ }ac\text{ }{{\left( c \right)}^{2}}~+\text{ }2a.\text{ }ac.\text{ }c \\ =\text{ }{{a}^{3}}c\text{ }+\text{ }a{{c}^{3}}~+\text{ }2{{a}^{2}}{{c}^{2}} \\ \end{array}\]

Taking ac as common

    \[\begin{array}{*{35}{l}} =\text{ }ac\text{ }({{a}^{2}}~+\text{ }{{c}^{2}}~+\text{ }2ac) \\ =\text{ }ac\text{ }{{\left( a\text{ }+\text{ }c \right)}^{2}} \\ \end{array}\]

RHS

    \[=\text{ }({{a}^{2}}~+\text{ }{{b}^{2}})\text{ }({{b}^{2}}~+\text{ }{{c}^{2}})\]

Using equation (1)

    \[=\text{ }({{a}^{2}}~+\text{ }ac)\text{ }(ac\text{ }+\text{ }{{c}^{2}})\]

Taking common terms out

    \[\begin{array}{*{35}{l}} =\text{ }a\text{ }\left( a\text{ }+\text{ }c \right)\text{ }c\text{ }\left( a\text{ }+\text{ }c \right) \\ =\text{ }ac\text{ }{{\left( a\text{ }+\text{ }c \right)}^{2}} \\ \end{array}\]

Hence, LHS = RHS.

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