# geometric and arithmetic sequence problem

• Apr 5th 2010, 09:37 AM
shawli
geometric and arithmetic sequence problem
An arithmetic series has a first term of $\displaystyle -4$ and a common difference of $\displaystyle 1$. A geometric series has a first term of $\displaystyle 8$ and a common ratio of $\displaystyle 0.5$. After how many terms does the sum of the arithmetic series exceed the sum of the geometric series?

I end up getting:

$\displaystyle \frac{n}{2}(n-9)>4(1-0.5^n)$

and I'm not sure how to solve it!

• Apr 5th 2010, 11:34 AM
HallsofIvy
So the arithmetic series is -4-3-2-1+0+1+2+... and the geometric series is $\displaystyle 8+ 4+ 2+ 1+ \frac{1}{2}+ \frac{1}{4}+ ...$.

The nth term in the arithmetic series is -4+ (n-1) and the sum up to that n is (-4+ -4+ n-1)n/2= n(n- 9)/2, exactly what you have.

The sum of the geometric series is $\displaystyle \frac{8(1- 0.5^{n-1})}{1- 0.5}= 16(1- 0.5^{n-1})$ which is almost what you have. Looks like you multiplied by 1/2 rather than dividing.

Now you want $\displaystyle n(n-9)/2> 16(1- 0.5^{n-1}$
Notice that the infinite serie would have sum 16. n(n-9)/2> 16 when $\displaystyle n(n- 9)= n^2- 9n> 32$ or $\displaystyle n^2- 9n- 32> 0$.

$\displaystyle n^2- 9n- 32= 0$ when $\displaystyle n= \frac{9\pm\sqrt{81+ 128}}{2}= \frac{9\pm\sqrt{209}}{2}$ which is about 11.7.
Now just check: when n= 11, the arithmetic sum is 11(11-9)/2= 11 and the geometric sum is $\displaystyle 16(1- .5^10)= 15.984375$. When n= 12, the arithmetic sum is 12(12- 9)/2= 18 and the geometric sum is [tex]16(1- .5^{11})= 15.9921875.
• Apr 5th 2010, 07:06 PM
shawli
Ohhhh I was supposed to use the fact that it was an infinite series there (and also not mess up my algebra).

Thank you!