Question

If \[{a^2},{b^2},{c^2}\] are in AP, then the following are also in AP: \[\dfrac{a}{{b + c}},\dfrac{b}{{c + a}},\dfrac{c}{{a + b}}\]
\[{\text{A}}{\text{. }}\]True
\[{\text{B}}{\text{. }}\]False

Answer 1

Hint- Here, we will be using the condition for three terms to be in an arithmetic progression (AP) i.e., three consecutive terms are said to be in arithmetic progression if twice of the middle term is equal to the sum of the other two remaining terms.

Complete step-by-step solution -
Let us suppose \[\dfrac{a}{{b + c}},\dfrac{b}{{c + a}},\dfrac{c}{{a + b}}\] are in AP.
As we know that if three terms are in AP then twice the middle term is equal to the sum of the other two terms.
i.e., \[{\text{2}}\left( {\dfrac{b}{{c + a}}} \right) = \dfrac{a}{{b + c}} + \dfrac{c}{{a + b}}\]
Now, if we achieve \[{\text{2}}\left( {\dfrac{b}{{c + a}}} \right) = \dfrac{a}{{b + c}} + \dfrac{c}{{a + b}}\], then we can say that our assumption is true.
\[
  \dfrac{b}{{c + a}} + \dfrac{b}{{c + a}} = \dfrac{a}{{b + c}} + \dfrac{c}{{a + b}} \\
   \Rightarrow \dfrac{b}{{c + a}} - \dfrac{a}{{b + c}} = \dfrac{c}{{a + b}} - \dfrac{b}{{c + a}} \\
 \]
Adding 1 and subtracting 1 on both sides of the above equation, we get
\[
   \Rightarrow \dfrac{b}{{c + a}} + 1 - \dfrac{a}{{b + c}} - 1 = \dfrac{c}{{a + b}} + 1 - \dfrac{b}{{c + a}} - 1 \\
   \Rightarrow \left[ {\dfrac{b}{{c + a}} + 1} \right] - \left[ {\dfrac{a}{{b + c}} + 1} \right] = \left[ {\dfrac{c}{{a + b}} + 1} \right] - \left[ {\dfrac{b}{{c + a}} + 1} \right] \\
   \Rightarrow \left[ {\dfrac{{b + c + a}}{{c + a}}} \right] - \left[ {\dfrac{{a + b + c}}{{b + c}}} \right] = \left[ {\dfrac{{c + a + b}}{{a + b}}} \right] - \left[ {\dfrac{{b + c + a}}{{c + a}}} \right] \\
 \]
Now, taking \[\left( {a + b + c} \right)\] common from both sides of the above equation we have
\[ \Rightarrow \left( {a + b + c} \right)\left\{ {\left[ {\dfrac{1}{{c + a}}} \right] - \left[ {\dfrac{1}{{b + c}}} \right]} \right\} = \left( {a + b + c} \right)\left\{ {\left[ {\dfrac{1}{{a + b}}} \right] - \left[ {\dfrac{1}{{c + a}}} \right]} \right\}\]
Cancelling \[\left( {a + b + c} \right)\] from both sides of the above equation, we get
\[ \Rightarrow \left[ {\dfrac{1}{{c + a}}} \right] - \left[ {\dfrac{1}{{b + c}}} \right] = \left[ {\dfrac{1}{{a + b}}} \right] - \left[ {\dfrac{1}{{c + a}}} \right]\]
Taking \[\left( {c + a} \right)\left( {b + c} \right)\] and \[\left( {c + a} \right)\left( {a + b} \right)\] as the LCM for the LHS and RHS respectively, we have
\[
   \Rightarrow \dfrac{{b + c - \left( {c + a} \right)}}{{\left( {c + a} \right)\left( {b + c} \right)}} = \dfrac{{c + a - \left( {a + b} \right)}}{{\left( {c + a} \right)\left( {a + b} \right)}} \\
   \Rightarrow \dfrac{{b + c - c - a}}{{\left( {b + c} \right)}} = \dfrac{{c + a - a - b}}{{\left( {a + b} \right)}} \\
   \Rightarrow \dfrac{{b - a}}{{\left( {b + c} \right)}} = \dfrac{{c - b}}{{\left( {a + b} \right)}} \\
 \]
By cross-multiplying the above equation, we get
\[
   \Rightarrow \left( {b - a} \right)\left( {b + a} \right) = \left( {c + b} \right)\left( {c - b} \right) \\
   \Rightarrow {b^2} - {a^2} = {c^2} - {b^2} \\
   \Rightarrow 2{b^2} = {a^2} + {c^2}{\text{ }} \to {\text{(1)}} \\
 \]
Now, if equation (1) holds true we can say that our assumption is true.
Since, it is given that \[{a^2},{b^2},{c^2}\] are in AP which means \[2{b^2} = {a^2} + {c^2}\]
Therefore, equation (1) holds true which means that our assumption is correct.
So, \[\dfrac{a}{{b + c}},\dfrac{b}{{c + a}},\dfrac{c}{{a + b}}\] are also in AP if \[{a^2},{b^2},{c^2}\] are in AP.
Hence, option A is correct i.e., True.

Note- In these type of problems, the condition for three terms to be in an AP is applied in order to find out whether the given three terms are in an AP or not, we simplify the equation which will come after applying the AP condition and if that equation holds true we can say that the given three terms are in AP.