Question

If a, b, c are in H.P. then find the value of $ \dfrac{b+a}{b-a}+\dfrac{b+c}{b-c} $ .
(a) 1
(b) 3
(c) 4
(d) 2

Answer 1

Hint: If a, b, c, are in H.P. The reciprocal of their individual terms will be in arithmetic progression that is A.P. and when an odd number of terms are given in A.P. then we should let the middle term of the series as reference for the other terms to simplify the equations. So we take the middle term as $ {{a}_{1}} $ and then accordingly we get the first and third term too. Then, again we take their reciprocals and substitute in the given expression to get the final answer.

Complete step-by-step answer:
Here it is given that a, b, c, are in H.P. hence, $ \dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c} $ are in A.P.
Now we have to express the terms of the A.P. in a way which will make our equation easy to simplify, whenever odd number of terms are given in an A.P. try to assume middle term of that A.P. as a random variable say $ {{a}_{1}} $ and other terms accordingly,
   $ \begin{align}
  & \dfrac{1}{a}={{a}_{1}}-d, \\
 & \dfrac{1}{b}={{a}_{1}}, \\
 & \dfrac{1}{c}={{a}_{1}}+d. \\
\end{align} $
Where d, is supposed to be the common difference of A.P. and $ {{a}_{1}} $ is any random variable equal to $ \dfrac{1}{b} $ .
After taking the reciprocal on both sides of above equation we can write them as,
\[\begin{align}
  & a=\dfrac{1}{{{a}_{1}}-d}, \\
 & b=\dfrac{1}{{{a}_{1}}}, \\
 & c=\dfrac{1}{{{a}_{1}}+d} \\
\end{align}\]
Now, put the above values of a, b, c as \[\dfrac{1}{{{a}_{1}}-d},\dfrac{1}{{{a}_{1}}},\dfrac{1}{{{a}_{1}}+d}\] in the given equation $ \dfrac{b+a}{b-a}+\dfrac{b+c}{b-c} $ .
And we’ll get the following result,
\[\dfrac{b+a}{b-a}+\dfrac{b+c}{b-c}=\dfrac{\dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{1}}-d}}{\dfrac{1}{{{a}_{1}}}-\dfrac{1}{{{a}_{1}}-d}}+\dfrac{\dfrac{1}{{{a}_{1}}}+\dfrac{1}{{{a}_{1}}+d}}{\dfrac{1}{{{a}_{1}}}-\dfrac{1}{{{a}_{1}}+d}}\]
Taking LCM of the terms in the numerator and denominator on the RHS, we get
\[\dfrac{b+a}{b-a}+\dfrac{b+c}{b-c}=\dfrac{\dfrac{{{a}_{1}}-d+{{a}_{1}}}{{{a}_{1}}({{a}_{1}}-d)}}{\dfrac{{{a}_{1}}-d-{{a}_{1}}}{{{a}_{1}}({{a}_{1}}-d)}}+\dfrac{\dfrac{{{a}_{1}}+d+{{a}_{1}}}{{{a}_{1}}({{a}_{1}}+d)}}{\dfrac{{{a}_{1}}+d-{{a}_{1}}}{{{a}_{1}}({{a}_{1}}+d)}}\]
Cancelling out similar terms and simplifying further on RHS, we will get
\[\dfrac{b+a}{b-a}+\dfrac{b+c}{b-c}=\dfrac{2{{a}_{1}}-d}{-d}+\dfrac{2{{a}_{1}}+d}{d}\]
\[\dfrac{b+a}{b-a}+\dfrac{b+c}{b-c}=\dfrac{1}{d}\left( -2{{a}_{1}}+d+2{{a}_{1}}+d \right)\]
\[\dfrac{b+a}{b-a}+\dfrac{b+c}{b-c}=\dfrac{1}{d}\times 2d\]
Cancelling out variable d, we get
\[\dfrac{b+a}{b-a}+\dfrac{b+c}{b-c}=2\]
Hence the value of the given expression $ \dfrac{b+a}{b-a}+\dfrac{b+c}{b-c} $ is 2.
So, the correct answer is “Option D”.

Note: To solve these types of problems one needs to know how to rearrange the terms during calculation. There are an odd number of terms in an A.P. in a given problem.There’s a lot of calculation so you also need to be attentive and patient while solving any problem of A.P. and always remember reciprocal of any H.P. always form an A.P.