Prove that $3^x,3^y$ and $3^z$ are successive terms of a G.P.

Question

Prove that $3^x,3^y$ and $3^z$ are successive terms of a geometric progression , if $x,y$ and $z$ are successive terms of an arithmetic progression.

How should this be proven? I have no clue

Answer 1

$x$, $y$, and $z$ are successive terms of an arithmetic progression. Consequently, there is an $s$ such that $y = x + s$ and $z = x+2s$.

We are to show $3^x$, $3^y = 3^{x+s} = 3^x 3^s$, $3^z = 3^{x+2s} = 3^{x+s+s} = 3^x 3^s 3^s$ is a geometric progression. Do you se how to finish?

Answer 2

If $x,y,z$ are succively in AP then let each of them be$a,a+d,a+2d$

Which means $$3^x,3^y,3^z=3^a,3^{a+d},3^{a+2d}$$

Which when written simply is $$3^a,3^d×3^a,3^{2d}×3^a$$

Which turns out to be the very definition of an GP.

Answer 3

Without loss of generality, assume $x < y< z.$ Since $x, y,$ and $z$ are in AP, we must have $ y-x = z-y.$ Therefore, $3^{y-x} = 3^{z-y}$. Hence, $(3^y)^2 = 3^x. 3^z$. So, we are done.

Answer 4

Let's prove a bit stronger statement.

Propostion Let $x_1, \dots, x_n\in \mathbb R$ and $0 < k \neq 1$ be another real number. Then numbers $$ k^{x_1}, \dots, k^{x_n}$$ are successive terms of geometric progression if and only if $x_1, \dots, x_n$ are successive terms of arithmetic progression.

Proof: Assume $k^{x_1}, \dots, k^{x_n}$ are successive terms of geometric progression, what means that there is a number $c\in \mathbb R$ such that $k^{x_{i+1}}=ck^{x_i}$ for $i=1, 2,\dots, n-1$. Observe that both sides are positive, so $c>0$, and as $k\neq 1$ we can take the logarithm to get $$ x_{i+1} = \log_k c + x_i$$ which are successive terms of arithmetic progression with difference $\log_kc$.

Now assume that $x_1, \dots, x_n$ are successive terms of an arithmetic progression with difference $a$, i.e. $x_{i+1}=x_i+a$ for $i=1, 2,\dots,n-1 $. Exponentiating with base $k$ we have $$k^{x_{i+1}} = k^{x_i+a}=k^{x_i}\cdot k^a$$ which are successive terms of a geometric progression with ratio $k^a$.