Question

If a, b, c is in A.P., \[\alpha ,\beta ,\gamma \] are in H.P., \[a\alpha ,b\beta ,c\gamma \] are in G.P.
Then prove that \[a:b:c = \dfrac{1}{\gamma }:\dfrac{1}{\beta }:\dfrac{1}{\alpha }\].

Answer 1

Hint:
We will find the value of \[b,\beta ,b\beta \] by using the properties of Arithmetic progression, Harmonic progression and geometric progression. Then, using the values of b and \[\beta \] in the equation of G.P. expression and then simplifying, we get the required answer.

Complete step by step solution:
Given that a, b, c is in A.P. Therefore,
 \[b = \dfrac{{a + c}}{2}\] …. (1)
Given that \[\alpha ,\beta ,\gamma \] are in H.P.
Therefore,
 \[\beta = \dfrac{{2\alpha \gamma }}{{\alpha + \gamma }}\] …. (2)
And \[a\alpha ,b\beta ,c\gamma \] are in G.P.
Therefore,
 \[{b^2}{\beta ^2} = ac\alpha \gamma \] …. (3)
Putting the values of equations (1) and (2) in equation (3), we get
 \[ \Rightarrow {\left( {\dfrac{{a + c}}{2}} \right)^2}{\left( {\dfrac{{2\alpha \gamma }}{{\alpha + \gamma }}} \right)^2} = ac\alpha \gamma \]
On simplification, we get
 \[ \Rightarrow \dfrac{{{{(a + c)}^2}}}{4} \times \dfrac{{4{\alpha ^2}{\gamma ^2}}}{{{{(\alpha + \gamma )}^2}}} = ac\alpha \gamma \]
Eliminating 4 from both numerator and denominator, we have
 \[ \Rightarrow \dfrac{{{{(a + c)}^2}}}{{ac}} \times \dfrac{{{\alpha ^2}{\gamma ^2}}}{{{{(\alpha + \gamma )}^2}}} = \alpha \gamma \]
On cross multiplication, we get
 \[ \Rightarrow \dfrac{{{{(a + c)}^2}}}{{ac}} = \dfrac{{\alpha \gamma }}{{{\alpha ^2}{\gamma ^2}}} \times {(\alpha + \gamma )^2}\]
Eliminating \[\alpha \gamma \] from both numerator and denominator.
 \[ \Rightarrow \dfrac{{{{(a + c)}^2}}}{{ac}} = \dfrac{{{{(\alpha + \gamma )}^2}}}{{\alpha \gamma }}\] …. (4)
Using the formula \[{(x + y)^2} = {x^2} + {y^2} + 2xy\] in equation (4) we get
 \[ \Rightarrow \dfrac{{{a^2} + {c^2} + 2ac}}{{ac}} = \dfrac{{{\alpha ^2} + 2\alpha \gamma + {\gamma ^2}}}{{\alpha \gamma }}\]
Taking the R.H.S. term to left side, we get
 \[ \Rightarrow \dfrac{{{a^2} + {c^2} + 2ac}}{{ac}} - \dfrac{{{\alpha ^2} + 2\alpha \gamma + {\gamma ^2}}}{{\alpha \gamma }} = 0\]
On simplification, we have
 \[ \Rightarrow \dfrac{a}{c} + \dfrac{c}{a} + 2 - \dfrac{\alpha }{\gamma } - 2 - \dfrac{\gamma }{\alpha } = 0\]
On rearranging we get
 \[ \Rightarrow \dfrac{a}{c} + \dfrac{c}{a} = \dfrac{\alpha }{\gamma } + \dfrac{\gamma }{\alpha }\]
Multiplying by \[\dfrac{a}{c}\] , we get
 \[ \Rightarrow {\left( {\dfrac{a}{c}} \right)^2} + 1 = \dfrac{a}{c}\left( {\dfrac{\alpha }{\gamma } + \dfrac{\gamma }{\alpha }} \right)\]
Taking R.H.S term to the left side, we get
 \[ \Rightarrow {\left( {\dfrac{a}{c}} \right)^2} - \dfrac{a}{c}\left( {\dfrac{\alpha }{\gamma } + \dfrac{\gamma }{\alpha }} \right) + 1 = 0\]
This can also be written as
 \[ \Rightarrow \left( {\dfrac{a}{c} - \dfrac{\alpha }{\gamma }} \right)\left( {\dfrac{a}{c} - \dfrac{\gamma }{\alpha }} \right) = 0\]
Here, we will take the first factor and we get
 \[ \Rightarrow \dfrac{a}{c} = \dfrac{\alpha }{\gamma }\]
On cross multiplication, we have
 \[ \Rightarrow a\gamma = c\alpha \] …. (5)
We can also write it as
 \[ \Rightarrow \dfrac{a}{{\dfrac{1}{\gamma }}} = \dfrac{c}{{\dfrac{1}{\alpha }}}\] …. (6)
Putting the value of equation (5) in equation (3), we get
 \[{b^2}{\beta ^2} = {a^2}{\gamma ^2}\]
Taking square root both sides, we get
 \[b\beta = a\gamma \].
i.e. \[\dfrac{b}{{\dfrac{1}{\beta }}} = \dfrac{a}{{\dfrac{1}{\gamma }}}\] …. (7)
Therefore, from equations (6) and (7), we get
 \[ \Rightarrow \dfrac{a}{{\dfrac{1}{\gamma }}} = \dfrac{b}{{\dfrac{1}{\beta }}} = \dfrac{c}{{\dfrac{1}{\alpha }}}\]

Hence,
 \[ \Rightarrow a:b:c = \dfrac{1}{\gamma }:\dfrac{1}{\beta }:\dfrac{1}{\alpha }\]

Note:
An arithmetic progression is a sequence of numbers in which each term is derived from the preceding term by adding or subtracting a fixed number called the common difference "d"
For example, the sequence 9, 6, 3, 0, -3, .... is an arithmetic progression with -3 as the common difference?
A geometric progression is a sequence in which each term is derived by multiplying or dividing the preceding term by a fixed number called the common ratio. For example, the sequence 4, -2, 1, \[ - \dfrac{1}{2}\],.... is a Geometric Progression (GP) for which \[ - \dfrac{1}{2}\] is the common ratio.
Harmonic Progression is defined as the series of real numbers which is calculated by taking reciprocals of Arithmetic progression which do not contain zero.