Find all sequences which are simultaneously an Arithmetic Progression and a Geometric Progression.
$\displaystyle a_n = a_1 + (n-1)d$ --- Arithmetic prog.
$\displaystyle a_n = a_1r^{n-1}$ --- Geometric prog.
from geometric prog.. we have $\displaystyle a_1 = \frac{a_n}{r^{n-1}}$
so, we have $\displaystyle a_n = \frac{a_n}{r^{n-1}} + (n-1)d \Longleftrightarrow a_nr^{n-1} = a_n + (n-1)dr^{n-1} \Longleftrightarrow a_n = \frac{(n-1)dr^{n-1}}{r^{n-1}-1}$
Hi
Let $\displaystyle (a_n)_{n\geq 0}$ be such a sequence. There exists two real numbers $\displaystyle d$ and $\displaystyle r$ such that for any non-negative integer $\displaystyle n$
$\displaystyle \begin{cases} a_{n+1}-a_n=d & (1)\\ a_{n+1}=ra_n & (2)\end{cases}$
Substitute (2) in (1) : $\displaystyle ra_n-a_n=d \,\,\,(3)$
- If $\displaystyle r\neq 1$ , $\displaystyle (3)\implies a_n=\frac{d}{r-1}$ hence $\displaystyle (a_n)$ is a constant sequence. This implies that the common difference has to be $\displaystyle 0$ hence $\displaystyle a_n=\frac{0}{r-1}=0$ for any non-negative integer $\displaystyle n$.
- If $\displaystyle r=1$ , $\displaystyle (3) \implies d=a_n-a_n=0$ hence $\displaystyle a_n=a_0$ for any non-negative integer $\displaystyle n$ and $\displaystyle (a_n)$ is a constant sequence.
At this point we've shown that if a progression is both arithmetic and geometric than the sequence is constant. Now, one has to show that if a sequence is constant then it is both arithmetic and geometric : it will allow us to claim that "A sequence is both arithmetic and geometric iff it is constant".