Question

If a,b,c are the 3 consecutive terms of an A.P. and x,y,z are 3 consecutive terms of a G.P. then the value of ${{x}^{b-c}}\times {{y}^{c-a}}\times {{z}^{a-b}}$ is?
a) 0
b) xyz
c) -1
d) 1

Answer 1

Hint:The common difference of AP series is constant. So, if a,b,c are in AP then, we can write, $a+c=2b$, or $b-a=c-b=d$, where d is the common difference between the terms. We will find the value of b-c and a-b individually and then put all the value in expression ${{x}^{b-c}}\times {{y}^{c-a}}\times {{z}^{a-b}}$ to get the correct answer.

Complete step-by-step answer:
It is given in the question that a,b,c are three consecutive terms of an AP, also x,y,z are the consecutive terms of GP, then we have to find the value of ${{x}^{b-c}}$, ${{y}^{c-a}}$, ${{z}^{a-b}}$.
As a,b,c are three consecutive terms of AP and known that the common difference between any two consecutive terms in AP is constant. So, let us assume that $aSo, from this AP property we can find the value of $(a-b)$, $(b-c)$ and $(c-a)$. Now, from the equation $b-a=c-b=d$, we get $(a-b)=-d$ and $(b-c)=-d$ and $(c-a)=2d$.
So, on putting the value of $(a-b)$, $(b-c)$ and $(c-a)$ in GP series, we get ${{x}^{-d}}\times {{y}^{2d}}\times {{z}^{-d}}$.
If we have three terms x,y,z in GP, then they have a common ratio between them, which can be found by dividing two consecutive terms. Now, from the properties of GP, we know that we can write $y=\sqrt{xz}$ because x,y,z are consecutive terms of GP.
Therefore we get- ${{x}^{-d}}\times {{(\sqrt{xz})}^{2d}}\times {{z}^{-d}}$.
We know that ${{a}^{n}}+{{b}^{n}}$ can be written as ${{(ab)}^{n}}$ , So, we get
= ${{x}^{-d}}\times {{z}^{-d}}\times {{(\sqrt{xz})}^{2d}}$
=${{(xz)}^{-d}}\times {{(xz)}^{\dfrac{1}{2}\times 2d}}$
= ${{(xz)}^{-d+d}}$
= 1
Therefore, the value of ${{x}^{b-c}}$, ${{y}^{c-a}}$, ${{z}^{a-b}}$ is 1, and hence the value of ${{x}^{b-c}}\times {{y}^{c-a}}\times {{z}^{a-b}}$ is 1.
Hence option d is correct.

Note: The common difference in AP is constant between any two consecutive terms of AP. So, if a,b,c,d,e are in AP then we can write (a-b) as –d as we know that (b-a) is d. Also we can write $c=a+2d,d=a+3d,e=a+4d$. Using this property of AP will reduce our effort to solve this question.