Where does the given arithmetic series meet the given geometric series?

The arithmetic series I have begins with 100, and increases in increments of 100. (so d=100, a1=100, a2=200, a3=300, etc)

The geometric series I have begins with 2 and increases in increments of 2. so r=2, a1=2, a2=4, a3=8, etc.)

I know the formulas an=d(n-1) + a1

and an=a1r^n-1

...but I don't know how to apply them to find at what term n the two series will meet.

Help, please? :)

Re: Where does the given arithmetic series meet the given geometric series?

Quote:

Originally Posted by

**GGC4** The arithmetic series I have begins with 100, and increases in increments of 100. (so d=100, a1=100, a2=200, a3=300, etc)

The geometric series I have begins with 2 and increases in increments of 2. so r=2, a1=2, a2=4, a3=8, etc.)

I know the formulas an=d(n-1) + a1

and an=a1r^n-1

...but I don't know how to apply them to find at what term n the two series will meet.

Help, please? :)

Let's call the first series $\displaystyle a_n$ and the second series $\displaystyle b_n$. We want to find $\displaystyle n$ such that $\displaystyle a_n=b_n$.

$\displaystyle a_n = d(n-1)+a_1 = 100(n-1)+100 = 100n - 100 + 100 = 100n$

$\displaystyle b_n = b_1r^{n-1} = 2\cdot 2^{n-1} = 2^{n-1+1} = 2^n $

Solve for $\displaystyle n$:

$\displaystyle \fbox{100n=2^n}$

Re: Where does the given arithmetic series meet the given geometric series?

Quote:

Originally Posted by

**GGC4** The arithmetic series I have begins with 100, and increases in increments of 100. (so d=100, a1=100, a2=200, a3=300, etc)

The geometric series I have begins with 2 and increases in increments of 2. so r=2, a1=2, a2=4, a3=8, etc.)

I know the formulas an=d(n-1) + a1

and an=a1r^n-1

...but I don't know how to apply them to find at what term n the two series will meet.

Well there is no solution to this particular question.

You would say $\displaystyle d(n-1)+a_1=ar^{n-1}$.

In this case $\displaystyle 100(n-1)+100=2(2)^{n-1}$. Which has no elementary solution.

Re: Where does the given arithmetic series meet the given geometric series?

I am very, very thankful to you both for your replies!!

I agree that the found equation does not look easy to solve, and am seriously questioning whether I wrote the problem down wrong. >_<

The 8 I wrote down looked kind of like a 6 (which would make the geometric series an arithmetic series), but the two equations would clearly never meet if one was increasing in increments of two and the other in increments of 1000. So, I guess I'll just have to wait it out till class on Friday.

Sorry for taking up your time, and again, THANK YOU^whatever the answer to this problem is.