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If a, b, c are in H.P, show that \[\dfrac{{a - b}}{{b - c}} = \dfrac{a}{c}\].

Last updated date: 13th Jul 2024
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Hint: Here in this question, we have to show the given equality. Given the harmonic progression sequence, as we know harmonic progression is the reciprocal of arithmetic progression by taking a common difference between two elements of sequence on further simplification by using arithmetic operations, we get the required solution.

Complete step-by-step answer:
A Harmonic Progression (HP) is defined as a sequence of real numbers which is determined by taking the reciprocals of the arithmetic progression (AP) that does not contain 0.
Given, a, b, c are in HP if \[\dfrac{1}{a}\], \[\dfrac{1}{b}\], \[\dfrac{1}{c}\] are in AP.
An arithmetic sequence or progression is defined as a sequence of numbers in which for every pair of consecutive terms, the second number is obtained by adding a fixed number to the first one.
The fixed number that must be added to any term of an AP to get the next term is known as the common difference of the AP.
Let us consider the sequence:
 \[\dfrac{1}{a}\], \[\dfrac{1}{b}\], \[\dfrac{1}{c}\].
Now take a common difference of above sequence i.e.,
\[ \Rightarrow \dfrac{1}{b} - \dfrac{1}{a} = \dfrac{1}{c} - \dfrac{1}{b}\]
Now take LCM ‘ab’ in LHS and ‘bc’ in RHS, then we have
\[ \Rightarrow \dfrac{{a - b}}{{ab}} = \dfrac{{b - c}}{{cb}}\]
Now taking cross multiplication, we get
\[ \Rightarrow \dfrac{{a - b}}{{b - c}} = \dfrac{{ab}}{{cb}}\]
On cancelling like terms in both numerator and denominator of RHS, we get
\[ \Rightarrow \dfrac{{a - b}}{{b - c}} = \dfrac{a}{c}\]
Hence proved
\[\therefore \dfrac{{a - b}}{{b - c}} = \dfrac{a}{c}\]

Note: Remember, in an arithmetic sequence, the common difference (an addition or subtraction) between any two consecutive terms of sequence is a constant or same. And should know the basic arithmetic operations like addition, multiplication, subtraction and division to simplify the sequence problems.