Question

If we are given, \[\dfrac{1}{{(b - a)}} + \dfrac{1}{{(b - c)}} = \left( {\dfrac{1}{a}} \right) + \left( {\dfrac{1}{c}} \right)\], then \[a,b,c\] are in
(1) \[AP\]
(2) \[GP\]
(3) \[HP\]
(4) none of these

Answer 1

Hint: we can find the correct solution for the given simple problem by simplifying the given expression if the given expression is of the form \[2b = a + c\] then it is in A.P. if it is of the form \[{b^2} = ac\] then it is in G.P. If it is of the form \[b = \dfrac{{c + a}}{{2ac}}\] then it is in A.P.

Complete step by step answer:
Now let us consider the given equation
\[\dfrac{1}{{(b - a)}} + \dfrac{1}{{(b - c)}} = \left( {\dfrac{1}{a}} \right) + \left( {\dfrac{1}{c}} \right)\]
Take LCM of the denominator on both LHS and RHS we get
\[\dfrac{{b - c + b - a}}{{(b - a)(b - c)}} = \dfrac{{c + a}}{{ac}}\]
We can rewrite the above expression as
\[ \Rightarrow \dfrac{{2b - (c + a)}}{{{b^2} - b(a + c) + ac}} = \dfrac{{c + a}}{{ac}}\]
On cross multiplication we get
\[ \Rightarrow 2bac - ac(a + c) = {b^2}(a + c) - b{(a + c)^2} + ac(a + c)\]
\[ \Rightarrow {b^2}(a + c) - b{(a + c)^2} + 2ac(a + c) - 2bac = 0\]
take \[b(a + c)\] as a common factor from first two terms and \[2ac\] from next two terms we get
\[ \Rightarrow b(a + c)(b - (a + c)) + 2ac(a + c - b) = 0\]
We can rewrite it as follows
\[ \Rightarrow b(a + c)(b - (a + c)) + 2ac(b - (a + c)) = 0\]
Again, taking common factor we get
\[ \Rightarrow [b(a + c) - 2ac](b - (a + c)) = 0\]
\[(b - (a + c)) \ne 0\]
\[b(a + c) - 2ac = 0\]
\[ \Rightarrow b = \dfrac{{2ac}}{{a + c}}\]
Therefore, a, b, c are in H.P.

So, the correct answer is “Option 3”.

Note: The condition for any three numbers to be in A.P is that the common difference should be the same between any two consecutive numbers. That is If we take any 3 numbers in A.P. Then it satisfies the condition \[2b = a + c\].
The condition for any three numbers to be in G.P is that the common ratios should be the same between any two consecutive numbers. That is If we take any 3 numbers in G.P. Then it satisfies the condition \[{b^2} = ac\].
 A Harmonic Progression (HP) is defined as a sequence of real numbers which is determined by taking the reciprocals of the arithmetic progression that does not contain 0. In harmonic progression, any term in the sequence is considered as the harmonic means. That is If we take any 3 numbers in H.P. Then it satisfies the condition \[\dfrac{1}{b} - \dfrac{1}{a} = \dfrac{1}{c} - \dfrac{1}{b}\]
 On simplification we get \[b = \dfrac{{c + a}}{{2ac}}\].