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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...

If ln⁡(a+c),  ln⁡(a−c),  ln⁡(a−2b+c)\ln (a+c),\,\,\ln (a-c),\,\,\ln (a-2b+c) are in A.P. then
1. a, b, ca,\,b,\,c are in A.P.
2. a2,  b2,  c2{{a}^{2}},\,\,{{b}^{2}},\,\,{{c}^{2}} are in A.P.
3. a, b, ca,\,b,\,c are in G.P.
4. a, b, ca,\,b,\,c are in H.P.

Explanation

Solution

In this question use the concept that if the numbers are in A.P. that is a, b, ca,\,b,\,c then the equation 2b=a+c2b=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)\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)d=\ln (a-c)-\ln (a+c)=\ln (a-2b+c)-\ln (a-c)
By using the logarithmic property that is ln⁡(MN)=ln⁡(M)−ln⁡(N)\ln \left( \dfrac{M}{N} \right)=\ln \left( M \right)-\ln \left( N \right) we get:
ln⁡((a−c)(a+c))=ln⁡((a−2b+c)(a−c))\ln \left( \dfrac{(a-c)}{(a+c)} \right)=\ln \left( \dfrac{(a-2b+c)}{(a-c)} \right)
On comparing we have,
(a−c)(a+c)=(a−2b+c)(a−c)\dfrac{(a-c)}{(a+c)}=\dfrac{(a-2b+c)}{(a-c)}
By cross multiplying we get:
(a−c)2=(a−2b+c)(a+c){{(a-c)}^{2}}=(a-2b+c)(a+c)
By using the property (a−b)2=a2−2ab+b2{{(a-b)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} we get:
a2−2ac+c2=(a−2b+c)(a+c){{a}^{2}}-2ac+{{c}^{2}}=(a-2b+c)(a+c)
By simplifying the two brackets we get:
a2−2ac+c2=a2−2ab+ac+ac−2bc+c2{{a}^{2}}-2ac+{{c}^{2}}={{a}^{2}}-2ab+ac+ac-2bc+{{c}^{2}}
By cancelling a2{{a}^{2}} and c2{{c}^{2}} on this equation and also further simplification we get:
−2ac=−2ab+ac+ac−2bc-2ac=-2ab+ac+ac-2bc
By rearranging the term we get:
−4ac=−2ab−2bc-4ac=-2ab-2bc
By dividing the 2 on both sides we get:
−2ac=−ab−bc-2ac=-ab-bc
Minus get cancelled on both side
2ac=ab+bc2ac=ab+bc
Divide abcabc on both sides we get:
2acabc=ababc+bcabc\dfrac{2ac}{abc}=\dfrac{ab}{abc}+\dfrac{bc}{abc}
By further simplification we get:
2b=1c+1a\dfrac{2}{b}=\dfrac{1}{c}+\dfrac{1}{a}
By rearranging the term we get:
2b=1a+1c\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, ca,\,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.