Question

If \[{a^{1/x}} = {b^{1/y}} = {c^{1/z}}\] and \[a,b,c\] are in G.P., then \[x,y,z\] will be in
A. A.P.
B. G.P.
C. H.P.
D. None of these

Answer 1

Hint:
We know that in G.P., there is a common ratio between the series' consecutive terms. So we can write \[b = ar\],\[c = br\], and \[c = a{r^2}\]. Enter these values into \[{a^{1/x}} = {b^{1/y}} = {c^{1/z}}\] and solve. Then simplify two terms to get the values in base ‘a’ and ‘r’ on either side. Then, using two different terms, solve them in the same way. Compare the outcomes and equate terms with the same base. Solve it to find out the answer.
Formula used:
The arithmetic progression is \[a,b,c\]
\[(b - a) = (c - b)\]
\[2b = a+c\]
The geometric progression is \[a,b,c\]
\[{b^2} = ac\].

Complete step-by-step solution:
We have been given in the question that \[{a^{1/x}} = {b^{1/y}} = {c^{1/z}}\]and\[a,b,c\] are in geometric progression.
Let us assume for the given expression
\[{a^{1/x}} = {b^{1/y}} = {c^{1/z}} = p\]
So, on solving for each term, it becomes
\[a = {p^x},b = {p^y},c = {p^z}\]
Since, we already know that \[a,b,c\] are in geometric progression.
Then,
\[ \Rightarrow {b^2} = ac\]--(1)
Now, putting the value of \[a,b,c\] in the equation (1), we get
\[{\left( {{p^y}} \right)^2} = \left( {{p^x}} \right)\left( {{p^z}} \right)\]
Apply exponent rules for the above equation:
When the bases are same, we have to add the powers according to the rules of multiplying exponents.
\[ \Rightarrow {p^{2y}} = {p^{(x + z)}}\]
Equating powers on both sides, we get
\[2y = x + z\]
So \[x,y\]and\[z\] are in Arithmetic Progression.
Therefore, If \[{a^{1/x}} = {b^{1/y}} = {c^{1/z}}\] and \[a,b,c\] are in G.P., then \[x,y,z\] will be in A.P
Hence, the correct option is (A).
Note:
Here, we have taken the first and seconds terms and first and third terms, we can easily prove the equation given in the question, because the calculations are easy. The equation we will get will become complex, if we chose some other terms and thus results in consuming more time in solving.