
If \[\dfrac{1}{{b - a}} + \dfrac{1}{{b - c}} = \dfrac{1}{a} + \dfrac{1}{c}\], then find the relation between \[a,b,c\].
A. AP
B. GP
C. HP
D. None of these
Answer
164.7k+ views
Hint First we will rewrite the equation in the form \[\dfrac{1}{{b - a}} - \dfrac{1}{c} = \dfrac{1}{a} - \dfrac{1}{{b - c}}\]. After that we will simplify both sides and establish a relation between \[a,b,c\]. From this relation we will decide whether \[a,b,c\] are in AP, or GP, or HP.
Formula used
If \[x,y,z\] are in AP, then \[\dfrac{1}{x}\],\[\dfrac{1}{y},\dfrac{1}{z}\] are in HP.
Complete step by step solution
Given equation is
\[\dfrac{1}{{b - a}} + \dfrac{1}{{b - c}} = \dfrac{1}{a} + \dfrac{1}{c}\]
Subtract both sides by \[\dfrac{1}{{b - c}}\] and \[\dfrac{1}{c}\]
\[\dfrac{1}{{b - a}} + \dfrac{1}{{b - c}} - \dfrac{1}{{b - c}} - \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{c} - \dfrac{1}{{b - c}} - \dfrac{1}{c}\]
\[ \Rightarrow \dfrac{1}{{b - a}} - \dfrac{1}{c} = \dfrac{1}{a} - \dfrac{1}{{b - c}}\]
Now simplify both sides
\[ \Rightarrow \dfrac{{c - b + a}}{{c\left( {b - a} \right)}} = \dfrac{{b - c - a}}{{a\left( {b - c} \right)}}\]
Taking common \[ - 1\] from right side
\[ \Rightarrow \dfrac{{c - b + a}}{{c\left( {b - a} \right)}} = - \dfrac{{ - b + c + a}}{{a\left( {b - c} \right)}}\]
Cancel out \[c - b + a\] from numerator of both sides
\[ \Rightarrow \dfrac{1}{{c\left( {b - a} \right)}} = - \dfrac{1}{{a\left( {b - c} \right)}}\]
Now apply the cross multiplication
\[ \Rightarrow a\left( {b - c} \right) = - c\left( {b - a} \right)\]
\[ \Rightarrow ab - ac = - bc + ac\]
Calculate the value of \[b\] in terms of \[a\] and \[c\]
\[ \Rightarrow ab + bc = 2ac\]
Taking common \[b\]from left side
\[ \Rightarrow b\left( {a + c} \right) = 2ac\]
\[ \Rightarrow b = \dfrac{{2ac}}{{a + c}}\]
Now taking the reciprocal of each sides
\[ \Rightarrow \dfrac{1}{b} = \dfrac{{a + c}}{{2ac}}\]
\[ \Rightarrow \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}\]
So \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\] are in Ap.
Thus \[a,b,c\] are in HP.
Hence the correct option is option C.
Note: To solve this question, you need to rewrite the equation as both sides contain at least two terms. Each side must contain all variables. In the equation, the variables are a,b, and c. So that we rearrange the equation \[\dfrac{1}{{b - a}} + \dfrac{1}{{b - c}} = \dfrac{1}{a} + \dfrac{1}{c}\] as \[\dfrac{1}{{b - a}} - \dfrac{1}{c} = \dfrac{1}{a} - \dfrac{1}{{b - c}}\]. Then simplify the equation. Remember if \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\] are in AP then a,b,c are in HP.
Formula used
If \[x,y,z\] are in AP, then \[\dfrac{1}{x}\],\[\dfrac{1}{y},\dfrac{1}{z}\] are in HP.
Complete step by step solution
Given equation is
\[\dfrac{1}{{b - a}} + \dfrac{1}{{b - c}} = \dfrac{1}{a} + \dfrac{1}{c}\]
Subtract both sides by \[\dfrac{1}{{b - c}}\] and \[\dfrac{1}{c}\]
\[\dfrac{1}{{b - a}} + \dfrac{1}{{b - c}} - \dfrac{1}{{b - c}} - \dfrac{1}{c} = \dfrac{1}{a} + \dfrac{1}{c} - \dfrac{1}{{b - c}} - \dfrac{1}{c}\]
\[ \Rightarrow \dfrac{1}{{b - a}} - \dfrac{1}{c} = \dfrac{1}{a} - \dfrac{1}{{b - c}}\]
Now simplify both sides
\[ \Rightarrow \dfrac{{c - b + a}}{{c\left( {b - a} \right)}} = \dfrac{{b - c - a}}{{a\left( {b - c} \right)}}\]
Taking common \[ - 1\] from right side
\[ \Rightarrow \dfrac{{c - b + a}}{{c\left( {b - a} \right)}} = - \dfrac{{ - b + c + a}}{{a\left( {b - c} \right)}}\]
Cancel out \[c - b + a\] from numerator of both sides
\[ \Rightarrow \dfrac{1}{{c\left( {b - a} \right)}} = - \dfrac{1}{{a\left( {b - c} \right)}}\]
Now apply the cross multiplication
\[ \Rightarrow a\left( {b - c} \right) = - c\left( {b - a} \right)\]
\[ \Rightarrow ab - ac = - bc + ac\]
Calculate the value of \[b\] in terms of \[a\] and \[c\]
\[ \Rightarrow ab + bc = 2ac\]
Taking common \[b\]from left side
\[ \Rightarrow b\left( {a + c} \right) = 2ac\]
\[ \Rightarrow b = \dfrac{{2ac}}{{a + c}}\]
Now taking the reciprocal of each sides
\[ \Rightarrow \dfrac{1}{b} = \dfrac{{a + c}}{{2ac}}\]
\[ \Rightarrow \dfrac{2}{b} = \dfrac{1}{a} + \dfrac{1}{c}\]
So \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\] are in Ap.
Thus \[a,b,c\] are in HP.
Hence the correct option is option C.
Note: To solve this question, you need to rewrite the equation as both sides contain at least two terms. Each side must contain all variables. In the equation, the variables are a,b, and c. So that we rearrange the equation \[\dfrac{1}{{b - a}} + \dfrac{1}{{b - c}} = \dfrac{1}{a} + \dfrac{1}{c}\] as \[\dfrac{1}{{b - a}} - \dfrac{1}{c} = \dfrac{1}{a} - \dfrac{1}{{b - c}}\]. Then simplify the equation. Remember if \[\dfrac{1}{a},\dfrac{1}{b},\dfrac{1}{c}\] are in AP then a,b,c are in HP.
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