# Let a, b, be positive real numbers. If $a,{{G}_{1}},{{G}_{2}},b$ are in geometric progression and $a,{{H}_{1}},{{H}_{2}},b$ are in harmonic progression show that

$\dfrac{{{G}_{1}}{{G}_{2}}}{{{H}_{1}}{{H}_{2}}}=\dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab}$

Last updated date: 25th Mar 2023

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Answer

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Hint: A geometric progression (GP), also called a geometric sequence, is a sequence of numbers that differ from each other by a common ratio. A harmonic progression is a sequence of real numbers formed by taking the reciprocals of an arithmetic progression.

Complete step-by-step answer:

It is given in the question that a, b are positive real numbers.

And it is also given that $a,{{G}_{1}},{{G}_{2}},b$ are in geometric progression, and we know that a geometric progression (GP), also called a geometric sequence, is a sequence of numbers that differ from each other by a common ratio.

Therefore, we can say that the ratio of two consecutive terms is the same.

\[\dfrac{{{G}_{1}}}{a}=\dfrac{b}{{{G}_{2}}}\]

Now we will cross-multiply the equation.

${{G}_{1}}{{G}_{2}}=ab....\left( i \right)$

Now it is also given in the question that $a,{{H}_{1}},{{H}_{2}},b$ are in the harmonic progression and we know that harmonic progression is a sequence of real numbers formed by taking the reciprocals of an arithmetic progression.

For example, $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}......$ series is a harmonic series, then according to it x, y, z…. is an A.P.

Since it is given in the question that $a,{{H}_{1}},{{H}_{2}},b$ are in harmonic progression, we can write that $\dfrac{1}{a},\dfrac{1}{{{H}_{1}}},\dfrac{1}{{{H}_{2}}},\dfrac{1}{b}$ are in A.P.

The first term of A.P is $\dfrac{1}{a}$ and let common differences be d . And we know that the general form of A.P is ${{a}_{n}}=a+\left( n-1 \right)d$ where a is the first term, the common difference is d and n is ${{n}^{th}}$ term.

Then we can say that

$\dfrac{1}{{{H}_{1}}}=\dfrac{1}{a}+d....\left( ii \right)$

And similarly,

$\dfrac{1}{{{H}_{2}}}=\dfrac{1}{a}+2d....\left( iii \right)$

And,

$\dfrac{1}{b}=\dfrac{1}{a}+3d....\left( iv \right)$

Now using equation (iii), we can find the value of d

$d=\dfrac{\dfrac{1}{b}-\dfrac{1}{a}}{3}=\dfrac{a-b}{3ab}....\left( v \right)$

Now substitute the value of d in equation (ii) using equation (v).

$\begin{align}

& \dfrac{1}{{{H}_{1}}}=\dfrac{1}{a}+\dfrac{a-b}{3ab} \\

& \dfrac{1}{{{H}_{1}}}=\dfrac{3b+a-b}{3ab} \\

\end{align}$

$\dfrac{1}{{{H}_{1}}}=\dfrac{a+2b}{3ab}$

Now, we take the reciprocal on both sides, we get,

${{H}_{1}}=\dfrac{3ab}{a+2b}....\left( vi \right)$

Now substitute the value of d in equation (iii) using equation (v).

$\begin{align}

& \dfrac{1}{{{H}_{2}}}=\dfrac{1}{a}+2\left( \dfrac{a-b}{3ab} \right) \\

& \dfrac{1}{{{H}_{2}}}=\dfrac{3b+2a-2b}{3ab} \\

& \dfrac{1}{{{H}_{2}}}=3\dfrac{2a+b}{ab} \\

\end{align}$

Now, we take the reciprocal on both sides, we get,

${{H}_{2}}=\dfrac{3ab}{2a+b}....\left( vii \right)$

Now we need to prove the equation $\dfrac{{{G}_{1}}{{G}_{2}}}{{{H}_{1}}{{H}_{2}}}=\dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab}$ that is given in the question. So, we will substitute the values on the left-hand side of the equation and check that it would be equal to the right-hand side of the equation or not.

Using equation (i), (vi) and (vii).

$\begin{align}

& \dfrac{{{G}_{1}}{{G}_{2}}}{{{H}_{1}}{{H}_{2}}}=\dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab} \\

& \dfrac{ab}{\left( \dfrac{3ab}{2b+a} \right)\left( \dfrac{3ab}{2a+b} \right)}=\dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab} \\

& \dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab}=\dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab} \\

\end{align}$

Hence, it proves that $\dfrac{{{G}_{1}}{{G}_{2}}}{{{H}_{1}}{{H}_{2}}}=\dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab}$.

Note: To solve the question we must write the terms ${{H}_{1}}\text{ and }{{H}_{2}}$ of a, b and avoid using any extra variable. Similarly, write ${{G}_{1}}\text{ and }{{G}_{2}}$ in terms of a, b. The possibility of mistake could be done here in understanding the significance of the Harmonic series. Remember we have to consider the reciprocal of its term as an arithmetic series.

Complete step-by-step answer:

It is given in the question that a, b are positive real numbers.

And it is also given that $a,{{G}_{1}},{{G}_{2}},b$ are in geometric progression, and we know that a geometric progression (GP), also called a geometric sequence, is a sequence of numbers that differ from each other by a common ratio.

Therefore, we can say that the ratio of two consecutive terms is the same.

\[\dfrac{{{G}_{1}}}{a}=\dfrac{b}{{{G}_{2}}}\]

Now we will cross-multiply the equation.

${{G}_{1}}{{G}_{2}}=ab....\left( i \right)$

Now it is also given in the question that $a,{{H}_{1}},{{H}_{2}},b$ are in the harmonic progression and we know that harmonic progression is a sequence of real numbers formed by taking the reciprocals of an arithmetic progression.

For example, $\dfrac{1}{x},\dfrac{1}{y},\dfrac{1}{z}......$ series is a harmonic series, then according to it x, y, z…. is an A.P.

Since it is given in the question that $a,{{H}_{1}},{{H}_{2}},b$ are in harmonic progression, we can write that $\dfrac{1}{a},\dfrac{1}{{{H}_{1}}},\dfrac{1}{{{H}_{2}}},\dfrac{1}{b}$ are in A.P.

The first term of A.P is $\dfrac{1}{a}$ and let common differences be d . And we know that the general form of A.P is ${{a}_{n}}=a+\left( n-1 \right)d$ where a is the first term, the common difference is d and n is ${{n}^{th}}$ term.

Then we can say that

$\dfrac{1}{{{H}_{1}}}=\dfrac{1}{a}+d....\left( ii \right)$

And similarly,

$\dfrac{1}{{{H}_{2}}}=\dfrac{1}{a}+2d....\left( iii \right)$

And,

$\dfrac{1}{b}=\dfrac{1}{a}+3d....\left( iv \right)$

Now using equation (iii), we can find the value of d

$d=\dfrac{\dfrac{1}{b}-\dfrac{1}{a}}{3}=\dfrac{a-b}{3ab}....\left( v \right)$

Now substitute the value of d in equation (ii) using equation (v).

$\begin{align}

& \dfrac{1}{{{H}_{1}}}=\dfrac{1}{a}+\dfrac{a-b}{3ab} \\

& \dfrac{1}{{{H}_{1}}}=\dfrac{3b+a-b}{3ab} \\

\end{align}$

$\dfrac{1}{{{H}_{1}}}=\dfrac{a+2b}{3ab}$

Now, we take the reciprocal on both sides, we get,

${{H}_{1}}=\dfrac{3ab}{a+2b}....\left( vi \right)$

Now substitute the value of d in equation (iii) using equation (v).

$\begin{align}

& \dfrac{1}{{{H}_{2}}}=\dfrac{1}{a}+2\left( \dfrac{a-b}{3ab} \right) \\

& \dfrac{1}{{{H}_{2}}}=\dfrac{3b+2a-2b}{3ab} \\

& \dfrac{1}{{{H}_{2}}}=3\dfrac{2a+b}{ab} \\

\end{align}$

Now, we take the reciprocal on both sides, we get,

${{H}_{2}}=\dfrac{3ab}{2a+b}....\left( vii \right)$

Now we need to prove the equation $\dfrac{{{G}_{1}}{{G}_{2}}}{{{H}_{1}}{{H}_{2}}}=\dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab}$ that is given in the question. So, we will substitute the values on the left-hand side of the equation and check that it would be equal to the right-hand side of the equation or not.

Using equation (i), (vi) and (vii).

$\begin{align}

& \dfrac{{{G}_{1}}{{G}_{2}}}{{{H}_{1}}{{H}_{2}}}=\dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab} \\

& \dfrac{ab}{\left( \dfrac{3ab}{2b+a} \right)\left( \dfrac{3ab}{2a+b} \right)}=\dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab} \\

& \dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab}=\dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab} \\

\end{align}$

Hence, it proves that $\dfrac{{{G}_{1}}{{G}_{2}}}{{{H}_{1}}{{H}_{2}}}=\dfrac{\left( 2a+b \right)\left( a+2b \right)}{9ab}$.

Note: To solve the question we must write the terms ${{H}_{1}}\text{ and }{{H}_{2}}$ of a, b and avoid using any extra variable. Similarly, write ${{G}_{1}}\text{ and }{{G}_{2}}$ in terms of a, b. The possibility of mistake could be done here in understanding the significance of the Harmonic series. Remember we have to consider the reciprocal of its term as an arithmetic series.

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