# Question on finite series?

• Jul 14th 2008, 06:27 AM
fardeen_gen
Question on finite series?
Find all sequences which are simultaneously an Arithmetic Progression and a Geometric Progression.
• Jul 14th 2008, 06:34 AM
kalagota
$a_n = a_1 + (n-1)d$ --- Arithmetic prog.
$a_n = a_1r^{n-1}$ --- Geometric prog.

from geometric prog.. we have $a_1 = \frac{a_n}{r^{n-1}}$

so, we have $a_n = \frac{a_n}{r^{n-1}} + (n-1)d \Longleftrightarrow a_nr^{n-1} = a_n + (n-1)dr^{n-1} \Longleftrightarrow a_n = \frac{(n-1)dr^{n-1}}{r^{n-1}-1}$
• Jul 14th 2008, 06:44 AM
fardeen_gen
I think for an Arithmetic Progression,
a(subscript n) = a(subscript 1) + (n - 1)d, isn't it?
• Jul 14th 2008, 06:49 AM
kalagota
oh yes.. just a typo error..
• Jul 14th 2008, 07:52 AM
flyingsquirrel
Hi
Quote:

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

Let $(a_n)_{n\geq 0}$ be such a sequence. There exists two real numbers $d$ and $r$ such that for any non-negative integer $n$

$\begin{cases} a_{n+1}-a_n=d & (1)\\ a_{n+1}=ra_n & (2)\end{cases}$

Substitute (2) in (1) : $ra_n-a_n=d \,\,\,(3)$

• If $r\neq 1$ , $(3)\implies a_n=\frac{d}{r-1}$ hence $(a_n)$ is a constant sequence. This implies that the common difference has to be $0$ hence $a_n=\frac{0}{r-1}=0$ for any non-negative integer $n$.
• If $r=1$ , $(3) \implies d=a_n-a_n=0$ hence $a_n=a_0$ for any non-negative integer $n$ and $(a_n)$ 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".