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Question

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(a) a, b and c are in H.P

(b) a, 2b and 3c are in A.P

(c) a, b and c are the sides of a triangle

(d) None of the above

Answer
Verified

Hint- In this question use the concept that if three numbers are in A.P that is a, b and c, then the equation $2b = a + c$ must hold true. This along with the basic logarithmic properties will help to approach the solution of this problem.

__Complete step-by-step solution -__

It is given that $\log a,\log b,\log c$ are in A.P

So according to the property of A.P common difference (d) should be equal.

$ \Rightarrow d = \log b - \log a = \log c - \log b$

$ \Rightarrow 2\log b = \log a + \log c$

Now using log property i.e. $x\log y = \log {y^x},\log m + \log n = \log mn$ so we have,

$ \Rightarrow \log {b^2} = \log ac$

Now on comparing we have,

$ \Rightarrow {b^2} = ac$

$ \Rightarrow \dfrac{b}{a} = \dfrac{c}{b}$ ................... (1)

Now it is also given that $\left( {\log a - \log 2b} \right),\left( {\log 2b - \log 3c} \right),\left( {\log 3c - \log a} \right)$ are in A.P.

So according to the property of A.P common difference (d) should be equal.

$ \Rightarrow d = \left[ {\left( {\log 2b - \log 3c} \right) - \left( {\log a - \log 2b} \right)} \right] = \left[ {\left( {\log 3c - \log a} \right) - \left( {\log 2b - \log 3c} \right)} \right]$

Now simplify the above equation we have,

$ \Rightarrow \left[ {2\log 2b - \log a - \log 3c} \right] = \left[ {2\log 3c - \log a - \log 2b} \right]$

$ \Rightarrow 3\log 2b = 3\log 3c$

Now divide by 3 throughout and on comparing we have,

$ \Rightarrow 2b = 3c$

$ \Rightarrow b = \dfrac{3}{2}c$

Now substitute this value in equation (1) we have,

$ \Rightarrow \dfrac{{\dfrac{3}{2}c}}{a} = \dfrac{c}{{\dfrac{3}{2}c}}$

Now simplify it we have,

$ \Rightarrow a = \dfrac{9}{4}c$

Therefore $a:b:c = \dfrac{9}{4}c:\dfrac{3}{2}c:c$

Now multiply by $\dfrac{4}{c}$ we have,

$ \Rightarrow a:b:c = 9:6:4$

Now as we all know to form a triangle the sum of any two sides must be greater than the third side.

So as we see in the calculated ratio the sum of any two sides is greater than the third side so a, b and c form a triangle.

So this is the required answer.

__Hence option (C) is the required answer.__

Note – It is always advised to remember the basic logarithmic identities like $\log {a^m} = m\log a$, ${\text{logmn = logm + logn}}$, $\log m - \log n = \log \dfrac{m}{n}$etc. A series is said to be in arithmetic progression if and only if the common difference that is the difference between the two consecutive terms always remains constant throughout the series.

It is given that $\log a,\log b,\log c$ are in A.P

So according to the property of A.P common difference (d) should be equal.

$ \Rightarrow d = \log b - \log a = \log c - \log b$

$ \Rightarrow 2\log b = \log a + \log c$

Now using log property i.e. $x\log y = \log {y^x},\log m + \log n = \log mn$ so we have,

$ \Rightarrow \log {b^2} = \log ac$

Now on comparing we have,

$ \Rightarrow {b^2} = ac$

$ \Rightarrow \dfrac{b}{a} = \dfrac{c}{b}$ ................... (1)

Now it is also given that $\left( {\log a - \log 2b} \right),\left( {\log 2b - \log 3c} \right),\left( {\log 3c - \log a} \right)$ are in A.P.

So according to the property of A.P common difference (d) should be equal.

$ \Rightarrow d = \left[ {\left( {\log 2b - \log 3c} \right) - \left( {\log a - \log 2b} \right)} \right] = \left[ {\left( {\log 3c - \log a} \right) - \left( {\log 2b - \log 3c} \right)} \right]$

Now simplify the above equation we have,

$ \Rightarrow \left[ {2\log 2b - \log a - \log 3c} \right] = \left[ {2\log 3c - \log a - \log 2b} \right]$

$ \Rightarrow 3\log 2b = 3\log 3c$

Now divide by 3 throughout and on comparing we have,

$ \Rightarrow 2b = 3c$

$ \Rightarrow b = \dfrac{3}{2}c$

Now substitute this value in equation (1) we have,

$ \Rightarrow \dfrac{{\dfrac{3}{2}c}}{a} = \dfrac{c}{{\dfrac{3}{2}c}}$

Now simplify it we have,

$ \Rightarrow a = \dfrac{9}{4}c$

Therefore $a:b:c = \dfrac{9}{4}c:\dfrac{3}{2}c:c$

Now multiply by $\dfrac{4}{c}$ we have,

$ \Rightarrow a:b:c = 9:6:4$

Now as we all know to form a triangle the sum of any two sides must be greater than the third side.

So as we see in the calculated ratio the sum of any two sides is greater than the third side so a, b and c form a triangle.

So this is the required answer.

Note – It is always advised to remember the basic logarithmic identities like $\log {a^m} = m\log a$, ${\text{logmn = logm + logn}}$, $\log m - \log n = \log \dfrac{m}{n}$etc. A series is said to be in arithmetic progression if and only if the common difference that is the difference between the two consecutive terms always remains constant throughout the series.

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