Question

 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 1

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.