Question

If \[\dfrac{{b + c - a}}{a}\], \[\dfrac{{c + a - b}}{b}\], \[\dfrac{{a + b - c}}{c}\] are in AP, then a, b, c are in
A. G.P
B. A.P
C. H.P
D. A.G.P

Answer 1

Hint: If any series is in A.P(Arithmetic Progression), then the difference between any two terms remains constant in the whole series. If a series \[{x_1}\], \[{x_2}\],\[{x_3}\]………….. \[{x_n}\] are in A.P, the difference between any two-term will be a constant (i.e. \[\begin{array}{*{20}{c}}
  {{x_1} - {x_1}}& = &{{x_3} - {x_2}}
\end{array}\]).

Formula Used:
If a, b, c are in AP then their reciprocals will be in HP.

Complete Step by step solution:
 We have to find that a, b, and c are in which series. For this purpose, we will have to convert the above terms into a, b, and c. To do this,
To simplify the above fractions, we will add 2 to each term of the series. So,
\[ \Rightarrow \dfrac{{b + c - a}}{a} + 2\], \[\dfrac{{c + a - b}}{b} + 2\], \[\dfrac{{a + b - c}}{c} + 2\] ……….. A.P(Arithmetic Progression)
and then
\[ \Rightarrow \dfrac{{b + c - a + 2a}}{a}\], \[\dfrac{{c + a - b + 2b}}{b}\], \[ \Rightarrow \dfrac{{a + b - c + 2c}}{c}\] ………………… A.P(Arithmetic Progression)
\[ \Rightarrow \dfrac{{a + b + c}}{a}\], \[\dfrac{{a + b + c}}{b}\], \[\dfrac{{a + b + c}}{c}\] ……………….. A.P(Arithmetic Progression)
Now, divide the above series by \[(a + b + c)\]
\[ \Rightarrow \dfrac{{(a + b + c)}}{{a(a + b + c)}}\], \[\dfrac{{(a + b + c)}}{{b(a + b + c)}}\], \[\dfrac{{(a + b + c)}}{{c(a + b + c)}}\] ………………. A.P(Arithmetic Progression)
\[ \Rightarrow \dfrac{1}{a}\], \[\dfrac{1}{b}\], \[\dfrac{1}{c}\] …………… A.P(Arithmetic Progression)
The above series are in A.P(Arithmetic Progression). As we know, the difference between any two terms in A.P remains constant.
So, we can write that
\[\begin{array}{*{20}{c}}
  { \Rightarrow {T_2} - {T_1}} = {{T_3} - {T_2}}
\end{array}\]
Therefore, we will get
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\dfrac{1}{b} - \dfrac{1}{a}}& = &{\dfrac{1}{c} - \dfrac{1}{b}}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  {\dfrac{{a - b}}{{ab}}}& = &{\dfrac{{b - c}}{{bc}}}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  {abc - {b^2}c}& = &{a{b^2} - abc}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  {2abc}& = &{a{b^2} + {b^2}c}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  {2abc}& = &{{b^2}(a + c)}
\end{array}\]
\[ \Rightarrow \begin{array}{*{20}{c}}
  b& = &{\dfrac{{2ac}}{{(a + c)}}}
\end{array}\]
This is the condition of H.P(Harmonic progression). Therefore, a, b, c are in H.P.

The correct option is C.

Note:The first point is to keep in mind that we will add such a number in each term to reduce the term into a, b, and c. If a,b and c are in A.P(Arithmetic progression), then the reciprocates of a, b, and c will be in H.P(Harmonic progression).