
If \[\dfrac{a}{{\left( {b + c} \right)}},{\rm{ }}\dfrac{b}{{\left( {c + a} \right)}},{\rm{ }}\dfrac{c}{{\left( {a + b} \right)}}\] are in AP, then?
A. \[a,b,c\] are in A.P
B. \[c,a,b\] are in A.P
C. \[{a^2},{\rm{ }}{b^2},{c^2}\] are in A.P
D. \[a,b,c\] are in G.P
Answer
164.1k+ views
Hint:
First we will apply the condition of AP on the given numbers. Then simplify the equation and establish a relation between \[a,b,c\]. From the relation we can conclude which option is the correct option.
Formula Used:
The difference between two consecutive terms is constant of an AP series.
d = b – a
Where, d - common difference
a,b - two consecutive terms.
Complete step-by-step answer:
Given \[\dfrac{a}{{\left( {b + c} \right)}},{\rm{ }}\dfrac{b}{{\left( {c + a} \right)}},{\rm{ }}\dfrac{c}{{\left( {a + b} \right)}}\]are in AP.
So second term – first term = third term – second term
\[\left[ {\dfrac{b}{{\left( {c + a} \right)}}} \right] - {\rm{ }}\left[ {\dfrac{a}{{\left( {b + c} \right)}}} \right]{\rm{ }} = {\rm{ }}\left[ {{\rm{ }}\dfrac{c}{{\left( {a + b} \right)}}} \right]{\rm{ }} - \left[ {\dfrac{b}{{\left( {c + a} \right)}}} \right]\]
Take LCM
\[({b^2} + bc - ac - {a^2})/\left( {b + c} \right)\left( {c + a} \right){\rm{ }} = {\rm{ }}({c^2} + ac - ab - {b^2})/\left( {a + b} \right)\left( {c + a} \right)\]
Cancel \[\left( {c + a} \right)\] on both sides and rearranging the numerator
\[\begin{array}{*{20}{l}}{({b^2} - {a^2} + bc - ac)/\left( {b + c} \right){\rm{ }} = {\rm{ }}({c^2} - {b^2} + ac - ab)/\left( {a + b} \right)}\\{\left( {\left( {b - a} \right)\left( {b + a} \right) + c\left( {b - a} \right)} \right)/\left( {b + c} \right){\rm{ }} = {\rm{ }}\left( {\left( {c - b} \right)\left( {c + b} \right) + a\left( {c - b} \right)} \right)/\left( {a + b} \right)}\end{array}\]
Taking( \[b-a\]) common on LHS and ( \[c - b\]) common on RHS
\[\begin{array}{*{20}{l}}{\dfrac{{\left( {b - a} \right)\left( {b + a + c} \right)}}{{\left( {b + c} \right)}}{\rm{ }} = {\rm{ }}\dfrac{{\left( {c - b} \right)\left( {c + b + a} \right)}}{{\left( {a + b} \right)}}}\\{\dfrac{{\left( {b - a} \right)}}{{\left( {b + c} \right)}} = \dfrac{{\left( {c - b} \right)}}{{\left( {a + b} \right)}}}\end{array}\]
By Cross Multiplication
\[\begin{array}{*{20}{l}}{(ba - {a^2} + {b^2} - ab){\rm{ }} = {\rm{ }}bc + {c^2} - {b^2} - bc}\\{{b^2} - {a^2}\; = {\rm{ }}{c^2} - {b^2}}\\{2{b^2}\; = {\rm{ }}{c^2} + {a^2}}\end{array}\]
Thus \[{a^2},{\rm{ }}{b^2},{c^2}\] are in AP.
Hence option C is the correct option.
Note:
Students are often confused with G.P and A.P. In A.P the difference between two consecutive terms is constant and in G.P the ratio of two consecutive terms is constant.
First we will apply the condition of AP on the given numbers. Then simplify the equation and establish a relation between \[a,b,c\]. From the relation we can conclude which option is the correct option.
Formula Used:
The difference between two consecutive terms is constant of an AP series.
d = b – a
Where, d - common difference
a,b - two consecutive terms.
Complete step-by-step answer:
Given \[\dfrac{a}{{\left( {b + c} \right)}},{\rm{ }}\dfrac{b}{{\left( {c + a} \right)}},{\rm{ }}\dfrac{c}{{\left( {a + b} \right)}}\]are in AP.
So second term – first term = third term – second term
\[\left[ {\dfrac{b}{{\left( {c + a} \right)}}} \right] - {\rm{ }}\left[ {\dfrac{a}{{\left( {b + c} \right)}}} \right]{\rm{ }} = {\rm{ }}\left[ {{\rm{ }}\dfrac{c}{{\left( {a + b} \right)}}} \right]{\rm{ }} - \left[ {\dfrac{b}{{\left( {c + a} \right)}}} \right]\]
Take LCM
\[({b^2} + bc - ac - {a^2})/\left( {b + c} \right)\left( {c + a} \right){\rm{ }} = {\rm{ }}({c^2} + ac - ab - {b^2})/\left( {a + b} \right)\left( {c + a} \right)\]
Cancel \[\left( {c + a} \right)\] on both sides and rearranging the numerator
\[\begin{array}{*{20}{l}}{({b^2} - {a^2} + bc - ac)/\left( {b + c} \right){\rm{ }} = {\rm{ }}({c^2} - {b^2} + ac - ab)/\left( {a + b} \right)}\\{\left( {\left( {b - a} \right)\left( {b + a} \right) + c\left( {b - a} \right)} \right)/\left( {b + c} \right){\rm{ }} = {\rm{ }}\left( {\left( {c - b} \right)\left( {c + b} \right) + a\left( {c - b} \right)} \right)/\left( {a + b} \right)}\end{array}\]
Taking( \[b-a\]) common on LHS and ( \[c - b\]) common on RHS
\[\begin{array}{*{20}{l}}{\dfrac{{\left( {b - a} \right)\left( {b + a + c} \right)}}{{\left( {b + c} \right)}}{\rm{ }} = {\rm{ }}\dfrac{{\left( {c - b} \right)\left( {c + b + a} \right)}}{{\left( {a + b} \right)}}}\\{\dfrac{{\left( {b - a} \right)}}{{\left( {b + c} \right)}} = \dfrac{{\left( {c - b} \right)}}{{\left( {a + b} \right)}}}\end{array}\]
By Cross Multiplication
\[\begin{array}{*{20}{l}}{(ba - {a^2} + {b^2} - ab){\rm{ }} = {\rm{ }}bc + {c^2} - {b^2} - bc}\\{{b^2} - {a^2}\; = {\rm{ }}{c^2} - {b^2}}\\{2{b^2}\; = {\rm{ }}{c^2} + {a^2}}\end{array}\]
Thus \[{a^2},{\rm{ }}{b^2},{c^2}\] are in AP.
Hence option C is the correct option.
Note:
Students are often confused with G.P and A.P. In A.P the difference between two consecutive terms is constant and in G.P the ratio of two consecutive terms is constant.
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