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Question: If a, b, c is in A.P., \[\alpha ,\beta ,\gamma \] are in H.P., \[a\alpha ,b\beta ,c\gamma \] are in ...

If a, b, c is in A.P., α,β,γ\alpha ,\beta ,\gamma are in H.P., aα,bβ,cγa\alpha ,b\beta ,c\gamma are in G.P.
Then prove that a:b:c=1γ:1β:1αa:b:c = \dfrac{1}{\gamma }:\dfrac{1}{\beta }:\dfrac{1}{\alpha }.

Explanation

Solution

We will find the value of b,β,bβb,\beta ,b\beta by using the properties of Arithmetic progression, Harmonic progression and geometric progression. Then, using the values of b and β\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=a+c2b = \dfrac{{a + c}}{2} …. (1)
Given that α,β,γ\alpha ,\beta ,\gamma are in H.P.
Therefore,
β=2αγα+γ\beta = \dfrac{{2\alpha \gamma }}{{\alpha + \gamma }} …. (2)
And aα,bβ,cγa\alpha ,b\beta ,c\gamma are in G.P.
Therefore,
b2β2=acαγ{b^2}{\beta ^2} = ac\alpha \gamma …. (3)
Putting the values of equations (1) and (2) in equation (3), we get
(a+c2)2(2αγα+γ)2=acαγ\Rightarrow {\left( {\dfrac{{a + c}}{2}} \right)^2}{\left( {\dfrac{{2\alpha \gamma }}{{\alpha + \gamma }}} \right)^2} = ac\alpha \gamma
On simplification, we get
(a+c)24×4α2γ2(α+γ)2=acαγ\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
(a+c)2ac×α2γ2(α+γ)2=αγ\Rightarrow \dfrac{{{{(a + c)}^2}}}{{ac}} \times \dfrac{{{\alpha ^2}{\gamma ^2}}}{{{{(\alpha + \gamma )}^2}}} = \alpha \gamma
On cross multiplication, we get
(a+c)2ac=αγα2γ2×(α+γ)2\Rightarrow \dfrac{{{{(a + c)}^2}}}{{ac}} = \dfrac{{\alpha \gamma }}{{{\alpha ^2}{\gamma ^2}}} \times {(\alpha + \gamma )^2}
Eliminating αγ\alpha \gamma from both numerator and denominator.
(a+c)2ac=(α+γ)2αγ\Rightarrow \dfrac{{{{(a + c)}^2}}}{{ac}} = \dfrac{{{{(\alpha + \gamma )}^2}}}{{\alpha \gamma }} …. (4)
Using the formula (x+y)2=x2+y2+2xy{(x + y)^2} = {x^2} + {y^2} + 2xy in equation (4) we get
a2+c2+2acac=α2+2αγ+γ2αγ\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
a2+c2+2acacα2+2αγ+γ2αγ=0\Rightarrow \dfrac{{{a^2} + {c^2} + 2ac}}{{ac}} - \dfrac{{{\alpha ^2} + 2\alpha \gamma + {\gamma ^2}}}{{\alpha \gamma }} = 0
On simplification, we have
ac+ca+2αγ2γα=0\Rightarrow \dfrac{a}{c} + \dfrac{c}{a} + 2 - \dfrac{\alpha }{\gamma } - 2 - \dfrac{\gamma }{\alpha } = 0
On rearranging we get
ac+ca=αγ+γα\Rightarrow \dfrac{a}{c} + \dfrac{c}{a} = \dfrac{\alpha }{\gamma } + \dfrac{\gamma }{\alpha }
Multiplying by ac\dfrac{a}{c} , we get
(ac)2+1=ac(αγ+γα)\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
(ac)2ac(αγ+γα)+1=0\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
(acαγ)(acγα)=0\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
ac=αγ\Rightarrow \dfrac{a}{c} = \dfrac{\alpha }{\gamma }
On cross multiplication, we have
aγ=cα\Rightarrow a\gamma = c\alpha …. (5)
We can also write it as
a1γ=c1α\Rightarrow \dfrac{a}{{\dfrac{1}{\gamma }}} = \dfrac{c}{{\dfrac{1}{\alpha }}} …. (6)
Putting the value of equation (5) in equation (3), we get
b2β2=a2γ2{b^2}{\beta ^2} = {a^2}{\gamma ^2}
Taking square root both sides, we get
bβ=aγb\beta = a\gamma .
i.e. b1β=a1γ\dfrac{b}{{\dfrac{1}{\beta }}} = \dfrac{a}{{\dfrac{1}{\gamma }}} …. (7)
Therefore, from equations (6) and (7), we get
a1γ=b1β=c1α\Rightarrow \dfrac{a}{{\dfrac{1}{\gamma }}} = \dfrac{b}{{\dfrac{1}{\beta }}} = \dfrac{c}{{\dfrac{1}{\alpha }}}

**Hence,
a:b:c=1γ:1β:1α\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, 12 - \dfrac{1}{2},.... is a Geometric Progression (GP) for which 12 - \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.