Question

If \[\dfrac{{b + a}}{{b - a}} = \dfrac{{b + c}}{{b - c}}\], then \[a,b,c\] are in
A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer 1

Hint: To understand the AP, GP, and HP. Take a note of the pattern the series is using. Check the pattern in the first four to five terms to learn what pattern the series is following after observing them, and then check the condition of all the series, including AP, GP, and HP. You can determine the type of series being given by carrying out this action.

Complete step by step solution: We have been given that,
\[\dfrac{{b + a}}{{b - a}} = \dfrac{{b + c}}{{b - c}}\]
Now, we have to cross multiply the given equation:
\[{b^2} - bc + ab - ac = {b^2} + bc - ab - ac\]
Subtract\[\left( {{b^2} - bc} \right)\]from both sides of the above equation, we get
\[{b^2} - bc + ab - ac - \left( {{b^2} - bc} \right) = {b^2} + bc - ab - ac - \left( {{b^2} - bc} \right)\]
Now, simplify the above equation, by canceling the similar terms, we get
\[{\rm{ab - ac = - ab + 2bc - ac}}\]
On subtracting\[\left( { - ab - ac} \right)\]from both sides of the above equation, we get
\[ab - ac - \left( { - ab - ac} \right) = - ab + 2bc - ac - \left( { - ab - ac} \right)\]
Again, simplify the above equation by canceling the similar terms, we obtain
\[{\rm{2ab = 2bc}}\]
On dividing both sides of the above equation by \[2b\], we get
\[\dfrac{{2ab}}{{2b}} = \dfrac{{2bc}}{{2b}}\]
Simplify the above equation by canceling the terms on numerator and denominator, we have
\[a = c\]
Therefore, if \[\dfrac{{b + a}}{{b - a}} = \dfrac{{b + c}}{{b - c}}\], then \[a = c\].

Hence, the option D is correct.

Note: A connection exists between Arithmetic progression, Geometric progression, and Harmonic Progression. Arithmetic progression is always greater than or equal to harmonic progression, according to the definition, we can conclude. If AP, GP, and HP mean a, b, and c, respectively, then\[{b^2} = ac\]is the relationship between them, and\[a > b > c\]is another relationship between their means. Students should be more careful in solving AP, GP and HP problems, because remembering formulas is a bit tough in this case.