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.