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If \[\frac{a}{b},\frac{b}{c},\frac{c}{a}\]are in H.P., then
A. \[{a^2}b,{c^2}a,{b^2}c,\] are in A.P.
B. \[{a^2}b,{c^2}a,{b^2}c,\] are in H.P.
C. \[{a^2}b,{c^2}a,{b^2}c,\] are in G.P.
D. None of these

Answer
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Hint
A mathematical sequence known as a geometric progression (GP) is one in which each following phrase is generated by multiplying each preceding term by a fixed integer, or "common ratio." This progression is sometimes referred to as a pattern-following geometric sequence of numbers. Here, each phrase is multiplied by the common ratio to generate the subsequent term, which is a non-zero value.
It should be remembered that the common ratio is obtained by dividing any next phrase by its preceding term. A sequence that results from multiplying or dividing each term of a GP by a non-zero constant is also a GP with the same common ratio.
Formula use:
The arithmetic progression is \[a,b,c\]
\[(b - a) = (c - b)\]
If \[a,b,c\] are in HP
\[\frac{1}{a} ,\frac{1}{b} ,\frac{1}{c}\] are in A.P
Complete step-by-step solution
The given equation is \[\frac{a}{b},\frac{b}{c},\frac{c}{a}\] is in the H.P series,
Then the reciprocal of this series be in A.P . that is,
\[\frac{c}{b} - \frac{b}{a} = \frac{a}{c} - \frac{c}{b}\]
\[ = > \frac{{(ac - {b^2})}}{{ab}} = \frac{{(ab - {c^2})}}{{bc}}\]
On both the sides, remove the value b and simplify the equation
\[c(ac - {b^2}) = a(ab - {c^2})\]
\[ = > {c^2}a - {b^2}c = {a^2}b - {c^2}a\]
By cancelling the similar terms, the equation becomes
\[ = > 2{c^2}a = {a^2}b + {b^2}c\]
Hence, the series \[{a^2}b,{c^2}a,{b^2}c,\] are in A.P.
Therefore, the correct option is A.

Note
The difference between any two consecutive integers in an arithmetic progression (AP) sequence of numbers is always the same amount. It also goes by the name Arithmetic Sequence. It is referred to as AP, a mathematical sequence in which there is always a constant difference between two terms. The common difference of the AP is the constant amount that must be multiplied by any phrase in order to obtain the subsequent term.