Question

If \[{{A}_{1}},{{A}_{2}};{{G}_{1}},{{G}_{2}}\] and \[{{H}_{1}},{{H}_{2}}\] be two A.Ms, G.Ms and H.Ms, between two quantities a and b then. Prove that, \[{{A}_{1}}{{H}_{2}}={{A}_{2}}{{H}_{1}}={{G}_{1}}{{G}_{2}}=ab\].

Answer 1

Hint:Find the equation of \[{{n}^{th}}\] term of AP, HP and GP. Thus find the values \[{{A}_{1}},{{A}_{2}},{{H}_{1}},{{H}_{2}},{{G}_{1}}\] and \[{{G}_{2}}\]. Thus find \[{{A}_{1}}{{H}_{2}},{{A}_{2}}H,{{G}_{1}}{{G}_{2}}\] and their value should be equal to ab.

Complete step-by-step answer:

From the equation \[{{A}_{1}},{{A}_{2}}\] are the A.Ms, \[{{H}_{1}}\] and \[{{H}_{2}}\] are the H.Ms and \[{{G}_{1}},{{G}_{2}}\] are \[G.Ms\], where A.M means arithmetic mean, G.M means geometric mean and H.M means harmonic mean.
We know that the \[{{n}^{th}}\] term of an AP is given by,
\[{{T}_{n}}=a+\left( n-1 \right)d\]
Where a is the first term and d is the common difference.
Similarly, the \[{{n}^{th}}\] term of a GP is given by
\[{{T}_{n}}=a{{r}^{n-1}}\], where r = common ratio.
We have been given two quantities a and b.
Hence, we can say that \[a,{{A}_{1}},{{A}_{2}},b\] are in AP.
Hence there are 4 terms i.e. n = 4.
\[{{T}_{n}}=a+\left( n-1 \right)d\]
Put, \[{{T}_{n}}=b\] and \[n=4\] in the above expression.
\[\begin{align}
  & \therefore b=a+\left( 4-1 \right)d \\
 & b=a+3d,a=b-3d \\
 & b-a=3d \\
 & d=\dfrac{b-a}{3} \\
\end{align}\]
Hence we got the common difference (d) of the AP as \[\left( \dfrac{b-a}{3} \right)\].
Now the first term here is a \[{{2}^{nd}}\] term is \[{{A}_{1}}\].
\[\begin{align}
  & \therefore {{A}_{1}}=a+d=a+\left( \dfrac{b-a}{3} \right) \\
 & {{A}_{1}}=\dfrac{3a+b-a}{3}=\dfrac{2a+b}{3} \\
\end{align}\]
Hence, we got the \[{{2}^{nd}}\] term, \[{{A}_{1}}=\dfrac{2a+b}{3}\].
Similarly, \[{{3}^{rd}}\] term, \[{{A}_{2}}=a+2d\]
\[\begin{align}
  & {{A}_{2}}=a+2\left( \dfrac{b-a}{3} \right)=\dfrac{3a+2b-2a}{3} \\
 & {{A}_{2}}=\dfrac{a+2b}{3} \\
\end{align}\]
\[\therefore \] The \[{{3}^{rd}}\] term of AP, \[{{A}_{2}}=\dfrac{a+2b}{3}\].
Hence, \[a,{{A}_{1}},{{A}_{2}},b\] are in AP.
\[\begin{align}
  & a=b-3d \\
 & {{A}_{1}}=\dfrac{2a+b}{3} \\
 & {{A}_{2}}=\dfrac{a+2b}{3} \\
\end{align}\]
Again we have \[a,{{G}_{1}},{{G}_{2}},b\] in GP.
\[{{T}_{n}}=a{{r}^{n-1}}\], put n = 4 and \[{{T}_{n}}=b\].
\[\begin{align}
  & \therefore b=a{{r}^{4-1}}=a{{r}^{3}} \\
 & b=a{{r}^{3}} \\
 & {{r}^{3}}=\dfrac{b}{a}\Rightarrow r={{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}} \\
\end{align}\]
Hence, we got the common difference of GP as \[{{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}}\].
Now, \[{{G}_{1}}=ar\], which is the \[{{2}^{nd}}\] term.
\[\begin{align}
  & {{G}_{1}}=a{{\left( \dfrac{b}{a} \right)}^{\dfrac{1}{3}}}=\dfrac{a}{{{a}^{\dfrac{1}{3}}}}.{{b}^{\dfrac{1}{3}}}=\left( {{a}^{1-\dfrac{1}{3}}} \right).{{b}^{\dfrac{1}{3}}} \\
 & {{G}_{1}}={{a}^{\dfrac{2}{3}}}.{{b}^{\dfrac{1}{3}}} \\
\end{align}\]
Hence, we got the \[{{2}^{nd}}\] term of GP as, \[{{G}_{1}}={{a}^{\dfrac{2}{3}}}.{{b}^{\dfrac{1}{3}}}\].
Similarly, \[{{3}^{rd}}\] term, \[{{G}_{2}}=a{{r}^{2}}\]
\[\begin{align}
  & {{G}_{2}}=a{{\left( \dfrac{b}{a} \right)}^{\dfrac{2}{3}}}=\dfrac{a}{{{a}^{\dfrac{2}{3}}}}.{{b}^{\dfrac{2}{3}}} \\
 & {{G}_{2}}=\left( {{a}^{1-\dfrac{2}{3}}} \right).{{b}^{\dfrac{2}{3}}}={{a}^{\dfrac{1}{3}}}.{{b}^{\dfrac{2}{3}}} \\
\end{align}\]
\[\therefore {{2}^{nd}}\] term of GP, \[{{G}_{2}}={{a}^{\dfrac{1}{3}}}.{{b}^{\dfrac{2}{3}}}\].
Hence, \[a,{{G}_{1}},{{G}_{2}},b\] are in GP.
\[\begin{align}
  & a=\dfrac{b}{{{r}^{3}}} \\
 & {{G}_{1}}={{a}^{\dfrac{2}{3}}}.{{b}^{\dfrac{1}{3}}} \\
 & {{G}_{2}}={{a}^{\dfrac{1}{3}}}.{{b}^{\dfrac{2}{3}}} \\
\end{align}\]
Now we have \[a,{{H}_{1}},{{H}_{2}},b\] in H.P.
Hence, HP \[={{\left( \dfrac{1}{n} \right)}^{th}}\] term of AP. Put, \[{{T}_{n}}=b\] and n = 4.
\[\begin{align}
  & \dfrac{1}{b}=\dfrac{1}{a}+\left( 4-1 \right)d \\
 & \dfrac{1}{b}=\dfrac{1}{a}+3d\Rightarrow 3d=\dfrac{1}{b}-\dfrac{1}{a}=\dfrac{a-b}{ab} \\
 & \therefore d=\dfrac{a-b}{3ab} \\
\end{align}\]
Now to get the \[{{2}^{nd}}\] term \[{{H}_{1}}\],
\[\begin{align}
  & \dfrac{1}{{{H}_{1}}}=\dfrac{1}{a}+d=\dfrac{1}{a}+\dfrac{a-b}{3ab} \\
 & \dfrac{1}{{{H}_{1}}}=\dfrac{1}{a}+\dfrac{1}{3b}-\dfrac{1}{3a} \\
 & \dfrac{1}{{{H}_{1}}}=\dfrac{3b+a-b}{3ab}=\dfrac{a+2b}{3ab} \\
 & \therefore {{H}_{1}}=\dfrac{3ab}{a+2b} \\
\end{align}\]
Similarly, \[{{3}^{rd}}\] term of H.M \[{{H}_{2}}\]
\[\begin{align}
  & \dfrac{1}{{{H}_{2}}}=\dfrac{1}{a}+2d=\dfrac{1}{a}+\dfrac{2}{3b}-\dfrac{2}{3a} \\
 & \dfrac{1}{{{H}_{2}}}=\dfrac{3b+2a-2b}{3ab}=\dfrac{b+2a}{3ab} \\
 & \therefore {{H}_{2}}=\dfrac{3ab}{2a+b} \\
\end{align}\]
Now let us find the value of \[{{A}_{1}}{{H}_{2}},{{A}_{2}}{{H}_{2}}\] and \[{{G}_{1}},{{G}_{2}}\].
\[\begin{align}
  & {{A}_{1}}{{H}_{2}}=\left( \dfrac{2a+b}{3} \right)\times \left( \dfrac{3ab}{2a+b} \right)=ab \\
 & {{A}_{2}}{{H}_{1}}=\left( \dfrac{a+2b}{3} \right)\times \left( \dfrac{3ab}{a+2b} \right)=ab \\
 & {{G}_{1}}{{G}_{2}}={{a}^{\dfrac{2}{3}}}{{b}^{\dfrac{1}{3}}}\times {{a}^{\dfrac{1}{3}}}{{b}^{\dfrac{2}{3}}}={{a}^{\left( \dfrac{2}{3}+\dfrac{1}{3} \right)}}.{{b}^{\left( \dfrac{1}{3}+\dfrac{2}{3} \right)}} \\
 & {{G}_{1}}{{G}_{2}}={{a}^{1}}.{{b}^{1}}=ab \\
\end{align}\]
Hence from the above, we can say that,
\[{{A}_{1}}{{H}_{2}}={{A}_{2}}{{H}_{1}}={{G}_{1}}{{G}_{2}}=ab\]
Hence, we have proved the expression.

Note: To solve a question like this you should know the basics of AP, GP and HP. Thus find the terms to prove the expression that have been given. Remember the formula to find the \[{{n}^{th}}\] term in AP, GP and HP.