Question

If \[\ln \left( a+c \right)\], \[\ln \left( c-a \right)\] and \[\ln \left( a-2b+c \right)\] are in A.P. then determine that elements \[a\], \[b\] and \[c\] are terms of which series.

(a) \[a\], \[b\] and \[c\] are in A.P.

(b) \[{{a}^{2}}\], \[{{b}^{2}}\] and \[{{c}^{2}}\] are in A.P.

(c) \[a\], \[b\] and \[c\] are in G.P.

(d) \[a\], \[b\] and \[c\] are in H.P.

Answer 1

Hint: In this question, We are given that \[\ln \left( a+c \right)\], \[\ln \left( c-a \right)\] and \[\ln \left( a-2b+c \right)\] are in A.P. Then there will be a common difference, say \[d\] between the consecutive terms of the A.P. Then determine an expression for \[a\], \[b\] and \[c\] by using properties of logarithms and hence find the relation between \[a\], \[b\] and \[c\].

Complete step-by-step answer:

We are given that \[\ln \left( a+c \right)\], \[\ln \left( c-a \right)\] and \[\ln \left( a-2b+c \right)\] are in A.P.

That is the elements \[\ln \left( a+c \right)\], \[\ln \left( c-a \right)\] and \[\ln \left( a-2b+c \right)\] are terms of an Arithmetic progression. Hence there will be a common difference between every consecutive term, say \[d\].

Then we have 

\[\ln \left( c-a \right)-\ln \left( a+c \right)=d..................(1)\]

And 

\[\ln \left( a-2b+c \right)-\ln \left( c-a \right)=d.............(2)\]

Now equating the value of the common difference \[d\] in equation (1) and equation (2), we get

\[\ln \left( a-2b+c \right)-\ln \left( c-a \right)=\ln \left( c-a \right)-\ln \left( a+c \right)\]

We will now take the terms \[\ln \left( c-a \right)\] on one side of the equation.

\[\ln \left( a-2b+c \right)+\ln \left( a+c \right)=2\ln \left( c-a \right)...............(3)\]

Now using the property of logarithm that \[\ln a+\ln b=\ln \left( ab \right)\], we will get

\[\ln \left( a-2b+c \right)+\ln \left( a+c \right)=\ln \left( \left( a-2b+c \right)\left( a+c \right) \right).............(4)\]

Using equation (4) in equation (3), we get

\[\ln \left( \left( a-2b+c \right)\left( a+c \right) \right)=2\ln \left( c-a \right)...............(5)\]

Again we will use the other property of logarithm, say \[a\ln x=\ln {{x}^{a}}\]. Thus we have 

 \[2\ln \left( c-a \right)=\ln {{\left( c-a \right)}^{2}}\]

Using the above value in equation (5), we get 

\[\ln \left( \left( a-2b+c \right)\left( a+c \right) \right)=\ln {{\left( c-a \right)}^{2}}...............(6)\]

Now we know the property that if \[\ln x=\ln y\], then \[x=y\].

Using the above property in equation (6), we get 

\[\left( a-2b+c \right)\left( a+c \right)={{\left( c-a \right)}^{2}}\]

On expanding the above equation we have

\[a\left( a-2b+c \right)+c\left( a-2b+c \right)={{\left( c-a \right)}^{2}}\]

\[\Rightarrow {{a}^{2}}-2ab+ac+ca-2bc+{{c}^{2}}={{c}^{2}}-2ac+{{a}^{2}}\]

Now on cancelling the equal terms of both sides of the above equation we get

\[-2ab+ac+ca-2bc=-2ac\]

\[\Rightarrow -2ab+2ac-2bc=-2ac\]

\[\Rightarrow 4ac=2ab+2bc\]

Dividing the above equation by 2, we get

\[2ac=ab+bc\]

Again of dividing the equation \[2ac=ab+bc\] by \[abc\] , we have

\[\dfrac{2ac}{abc}=\dfrac{ab}{abc}+\dfrac{bc}{abc}\]

\[\Rightarrow \dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}\]

Hence \[a\], \[b\] and \[c\] are in H.P.

So, the correct answer is “Option D”.

Note: In this problem, we will use required properties of logarithm. Also keep in mind the properties of terms in A.P, G.P and H.P. Then check which property is being satisfied by the terms \[a\], \[b\] and \[c\] to choose the correct answer.