Q.If $a,b,c$ are in geometric progression, and if $a^{\frac1x}=b^{\frac1y}=c^{\frac1z}$, then prove that $x,y,z$ are in arithmetic progression.

A sequence is an ordered list of numbers; formally, a finite sequence on $n$ terms is a function $f:\{1,2,\ldots,n\}\to\mathbb R$ with $f(k)=a_k$, while an infinite sequence is a function on all of $\mathbb N$. Terms may repeat (unlike the elements of a set).

Arithmetic progression (AP): $a,\,a+d,\,a+2d,\ldots$, where $d$ is the common difference. The $n^{th}$ term is

Every term (after the first) is the arithmetic mean of its neighbours: $a_k=\dfrac{a_{k-1}+a_{k+1}}2$ (Result 5.1) — equivalently, $T_{m+n}+T_{m-n}=2T_m$ for any valid $m,n$.

Geometric progression (GP): $a,\,ar,\,ar^2,\ldots$ ($a\ne0,r\ne0$), where $r$ is the common ratio. The $n^{th}$ term is

Every term is the geometric mean of its neighbours: $a_k=\sqrt{a_{k-1}a_{k+1}}$ (Result 5.2), so $t_{n-k},t_n,t_{n+k}$ is again a GP for any $k$. Taking logs of a GP with $r>0$ turns it into an AP with common difference $\log r$.

Arithmetico-geometric progression (AGP): term-by-term product of an AP and a GP, $a,(a+d)r,(a+2d)r^2,\ldots$, with $n^{th}$ term $T_n=(a+(n-1)d)r^{n-1}$. Setting $r=1$ recovers an AP; setting $d=0$ recovers a GP — AP and GP are special cases of AGP.

Harmonic progression (HP): a sequence whose reciprocals form an AP: $h_1,h_2,\ldots$ is an HP exactly when $\dfrac1{h_1},\dfrac1{h_2},\ldots$ is an AP, giving the general HP form $\dfrac1a,\dfrac1{a+d},\dfrac1{a+2d},\ldots$ (provided no denominator vanishes). If $a,b,c$ are in HP then $b=\dfrac{2ac}{a+c}$. …