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Question: If a, b, c are in A.P. then \(\dfrac{a}{{bc}},\,\dfrac{1}{c},\,\dfrac{1}{b}\) are in A) A.P. B) ...

If a, b, c are in A.P. then abc,1c,1b\dfrac{a}{{bc}},\,\dfrac{1}{c},\,\dfrac{1}{b} are in
A) A.P.
B) G.P.
C) H.P.
D) None of these

Explanation

Solution

In order to solve this question, you have to know about A.P. In this question a, b, c are in A.P. which means the double of the second term is the addition of the first term and third term i.e. 2b=a+c2b = a + c.
Now, just convert our given equation abc,1c,1b\dfrac{a}{{bc}},\,\dfrac{1}{c},\,\dfrac{1}{b} like the general equation of A.P. and you will get the correct answer.

Complete step by step answer:
We are given that the variables a, b, c are in arithmetic progression.
We know that when variables are in arithmetic progression then this means the second term is the addition of the first term and third term.
Here the first term is aa, the second term is bb and the third term is cc.
Which means 2b=a+c2b = a + c .
And we have found abc,1c,1b\dfrac{a}{{bc}},\,\dfrac{1}{c},\,\dfrac{1}{b} are in which progress.
2b=a+c\Rightarrow 2b = a + c
Now, we have to divide by bcbc on both the side of above equation and we will find our new equation looks similar to general equation of arithmetic progression,
Divide by bcbc
2c=abc+1b\Rightarrow \dfrac{2}{c} = \dfrac{a}{{bc}} + \dfrac{1}{b}
Now, take least common multiple (L.C.M.) of above equation and we will get,
2c=a+cbc\Rightarrow \dfrac{2}{c} = \dfrac{{a + c}}{{bc}}
Let’s cancel cc from both the side and we will get,
2=a+cb\Rightarrow 2 = \dfrac{{a + c}}{b}
Take bb to the left side and finally we will see beloved equation,
2b=a+c\Rightarrow 2b = a + c
See this is the equation of arithmetic progression.
Therefore, we can clearly see that abc,1c,1b\dfrac{a}{{bc}},\,\dfrac{1}{c},\,\dfrac{1}{b} are in A.P.
So, the correct option is (A).

Note:
Relation between arithmetic progression (A.P.), geometric progression (G.P.) and harmonic progression (H.P.): If A.M denotes the arithmetic mean, G.M denotes the geometric mean and H.M, the harmonic mean, then the relationship between the three is given by,
A.M.×G.M.=(H.M.)2A.M. \times G.M. = {(H.M.)^2}