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Question: If \[\dfrac{a+b}{1-ab},b,\dfrac{b+c}{1-bc}\] are in AP the \[a,\dfrac{1}{b},c\] are in 1\. AP 2\...

If a+b1−ab,b,b+c1−bc\dfrac{a+b}{1-ab},b,\dfrac{b+c}{1-bc} are in AP the a,1b,ca,\dfrac{1}{b},c are in
1. AP
2. GP
3. HP
4. None of these

Explanation

Solution

Here we have been given three variables that are in AP (Arithmetic Progression) . We have to find the progression in which the other three values given are. Firstly, we will use the formula when three numbers are in AP on the first set of values and simplify them till we get the relation between the three variables. Then we will rearrange the value we get so that it forms a progression and get our desired answer.

Complete answer:
The AP is given as follows,
a+b1−ab,b,b+c1−bc\dfrac{a+b}{1-ab},b,\dfrac{b+c}{1-bc}…..(1)\left( 1 \right)
We have to find the progression in which the below values are,
a,1b,ca,\dfrac{1}{b},c…..(2)\left( 2 \right)
Now as we know that if the values are in AP the difference between the consecutive term is equal using this concept in equation (1) we get,
b−a+b1−ab=b+c1−bc−bb-\dfrac{a+b}{1-ab}=\dfrac{b+c}{1-bc}-b
We will take LCM on both sides and simplify it as follows,
⇒b(1−ab)−(a+b)1−ab=b+c−b(1−bc)1−bc\Rightarrow \dfrac{b\left( 1-ab \right)-\left( a+b \right)}{1-ab}=\dfrac{b+c-b\left( 1-bc \right)}{1-bc}
⇒b−ab2−a−b1−ab=b+c−b+b2c1−bc\Rightarrow \dfrac{b-a{{b}^{2}}-a-b}{1-ab}=\dfrac{b+c-b+{{b}^{2}}c}{1-bc}
Cancelling the common terms we get,
⇒ab2−a1−ab=c+b2c1−bc\Rightarrow \dfrac{a{{b}^{2}}-a}{1-ab}=\dfrac{c+{{b}^{2}}c}{1-bc}
Taking common terms in both numerators we get,
⇒−a(b2+1)1−ab=c(1+b2)1−bc\Rightarrow \dfrac{-a\left( {{b}^{2}}+1 \right)}{1-ab}=\dfrac{c\left( 1+{{b}^{2}} \right)}{1-bc}
Common terms get cancel out from both sides,
⇒−a1−ab=c1−bc\Rightarrow \dfrac{-a}{1-ab}=\dfrac{c}{1-bc}
⇒−a(1−bc)=c(1−ab)\Rightarrow -a\left( 1-bc \right)=c\left( 1-ab \right)
Simplifying further we get,
⇒−a+abc=c−abc\Rightarrow -a+abc=c-abc
⇒abc+abc=a+c\Rightarrow abc+abc=a+c
So we get,
2abc=a+c2abc=a+c
Now multiply the above value by acac we get,
⇒2abcac=aac+cac\Rightarrow \dfrac{2abc}{ac}=\dfrac{a}{ac}+\dfrac{c}{ac}
⇒2b=1c+1a\Rightarrow 2b=\dfrac{1}{c}+\dfrac{1}{a}….(3)\left( 3 \right)
We can see that if equation (2) terms can’t be in GP as there is no such term which when multiply by the first value give us the second value so we will check for HP.
Now as know that for terms to be in HP the difference between the reciprocal of consecutive terms is the same.
So we have the terms as a,1b,ca,\dfrac{1}{b},c
So there reciprocal will be 1a,b,1c\dfrac{1}{a},b,\dfrac{1}{c} using HP concept we get,
⇒b−1a=1c−b\Rightarrow b-\dfrac{1}{a}=\dfrac{1}{c}-b
⇒b+b=1a+1c\Rightarrow b+b=\dfrac{1}{a}+\dfrac{1}{c}
So we get,
2b=1a+1b2b=\dfrac{1}{a}+\dfrac{1}{b}….(4)\left( 4 \right)
So values in equation (3) and (4) are the same so the values a,1b,ca,\dfrac{1}{b},c are in HP.
Hence the correct answer is (3).

Note:
In this type of question usually solving the terms whose progression is given by the concept of the particular progression we get the relation or progression of the terms we are finding. The concept of AP (Arithmetic Progression), GP (geometric Progression) and HP (Harmonic Progression) should be clear in order to solve such questions. Harmonic progression has the same concept as Arithmetic progression; the only difference is that in HP we find the difference between the reciprocal values.