# Thread: Question on finite series?

1. ## Question on finite series?

Find all sequences which are simultaneously an Arithmetic Progression and a Geometric Progression.

2. $\displaystyle a_n = a_1 + (n-1)d$ --- 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 $\displaystyle a_n = \frac{a_n}{r^{n-1}} + (n-1)d \Longleftrightarrow a_nr^{n-1} = a_n + (n-1)dr^{n-1} \Longleftrightarrow a_n = \frac{(n-1)dr^{n-1}}{r^{n-1}-1}$

3. --- Arithmetic prog.
I think for an Arithmetic Progression,
a(subscript n) = a(subscript 1) + (n - 1)d, isn't it?

4. oh yes.. just a typo error..

5. Hi
Originally Posted by fardeen_gen
Find all sequences which are simultaneously an Arithmetic Progression and a Geometric Progression.
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".