ISI2017-DCG-8

Question

If $x,y,z$ are in $A.P.$ and $a>1$, then $a^x, a^y, a^z$ are in

• A. $A.P.$
• B. $G.P$
• C. $H.P$
• D. none of these

Answer 1

Answer: $\mathbf B$

Solution:

Let $a = 2$

and, $\mathrm{x = 4, y = 6, z = 8}$ be the terms in $\mathbf {AP}$ with common difference $=2$

Now,

$\mathrm{a^x = 4^2, a^y = 4^4,  a^z = 4^6}$

$\mathrm {\frac{a^y}{a^x} = \frac{4^4}{4^2} = 16,\;\text{and} \;\frac{a^z}{a^y} = \frac{4^6}{4^4} = 16}$

So, the resultant terms are in $\mathbf GP$

$\therefore \mathbf B$ is the correct option.

Answer 2

It is given x,y,z are in AP.

so we know that from the property of AP,

2y=(x+z)

or,y=(x+z)/2

now we know that if a,b,c are in GP then,

$b^{2}$=ac

now it is given ,$a^{x},a^{y},a^{z}$

now we put the value of y in $a^{y}$ =$a^{(x+z)/2}$

now multiply $a^{x} ,a^{z}$ we get $a^{x+z}$

now if we compare it with the GP form we get ,

$(a^{^{y}})^{2}$=$a^{x+z}$

so $(a^{^{y}})^{2}$ =$a^{x} .a^{z}$

so it is in G.P.

Answer 3

It is given that x, y, z are in A.P so,

2y = x+z

Taking them powers to a on both sides, we get,

$a^{2y} = a^{{x+z}}$

$a^{y}. a^{y} = a^{x} .a^{z}$

Rearranging the terms, we get,

$\frac{a^{y}}{a^{x}} = \frac{a^{z}}{a^{y}}$

which is the condition for G.P.

Hence we get $a^{x}$, $a^{y}$, $a^{z}$ are in G.P.