Question

If \[\ln (a+c),\,\,\ln (a-c),\,\,\ln (a-2b+c)\] are in A.P. then
1. \[a,\,b,\,c\] are in A.P.
2. \[{{a}^{2}},\,\,{{b}^{2}},\,\,{{c}^{2}}\] are in A.P.
3. \[a,\,b,\,c\] are in G.P.
4. \[a,\,b,\,c\] are in H.P.

Answer 1

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

Complete step by step answer:
It is given that \[\ln (a+c),\,\,\ln (a-c),\,\,\ln (a-2b+c)\] are in A.P.
So according to the property of A.P Common difference (d) should be equal
\[d=\ln (a-c)-\ln (a+c)=\ln (a-2b+c)-\ln (a-c)\]
By using the logarithmic property that is \[\ln \left( \dfrac{M}{N} \right)=\ln \left( M \right)-\ln \left( N \right)\] we get:
\[\ln \left( \dfrac{(a-c)}{(a+c)} \right)=\ln \left( \dfrac{(a-2b+c)}{(a-c)} \right)\]
On comparing we have,
\[\dfrac{(a-c)}{(a+c)}=\dfrac{(a-2b+c)}{(a-c)}\]
By cross multiplying we get:
\[{{(a-c)}^{2}}=(a-2b+c)(a+c)\]
By using the property \[{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\] we get:
\[{{a}^{2}}-2ac+{{c}^{2}}=(a-2b+c)(a+c)\]
By simplifying the two brackets we get:
\[{{a}^{2}}-2ac+{{c}^{2}}={{a}^{2}}-2ab+ac+ac-2bc+{{c}^{2}}\]
By cancelling \[{{a}^{2}}\] and \[{{c}^{2}}\] on this equation and also further simplification we get:
\[-2ac=-2ab+ac+ac-2bc\]
By rearranging the term we get:
\[-4ac=-2ab-2bc\]
By dividing the 2 on both sides we get:
\[-2ac=-ab-bc\]
Minus get cancelled on both side
\[2ac=ab+bc\]
Divide \[abc\] on both sides we get:
\[\dfrac{2ac}{abc}=\dfrac{ab}{abc}+\dfrac{bc}{abc}\]
By further simplification we get:
\[\dfrac{2}{b}=\dfrac{1}{c}+\dfrac{1}{a}\]
By rearranging the term we get:
\[\dfrac{2}{b}=\dfrac{1}{a}+\dfrac{1}{c}\]
This above equation looks like a general formula for harmonic progression that means it is a reciprocal of Arithmetic progression.
So, \[a,\,b,\,c\] are in H.P.

So, the correct answer is “Option 4”.

Note: It is always advisable to remember the basic logarithmic identity. 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. So, in this way the above solution is referred for solving such types of problems.