How Do You Solve Problems Involving Geometric Sequences and Series?

[Astronomer107]

I'm trying to get an A in honors AlgII/Trig and it is impossible, but I won't give up, so I have a few questions.

I'm not sure how to find the first two terms of a sequence (I got a few right, but most wrong and I don't know what's wrong). One of the problems is: a5 = 20; a8 = 4/25.

I set the problem up in this manner: 4/25 = -20r^3, but it does not turn out to be a perfect cube and the answer has no radicals in it, so it must work somehow.

Also: a1=2, a2=5, an=625/8, n=? I tried to find the common ratio, which I though was 2/5 (unless I forgot my basic algebra??), but it the whole problem is messed up.

Only one more thing, I promise! I'm pretty much altogether confused on geometric series. I was given the formula:

Sn= (a1-a1r^n)/1-r where r= common ratio and n is the number of terms in the series. One of the problems was: find the sum of the series: a1= 3/5, r=2, n-7. When I did this, it turned out to be a horrible number, when in fact the answer was supposed to be 47 and 5/8. If anyone could please just explain some of these things to me (I was making up a quiz while the teacher was explaining it). Thank you so much!

 

[quantumdude]

Originally posted by Astronomer107
I'm not sure how to find the first two terms of a sequence (I got a few right, but most wrong and I don't know what's wrong). One of the problems is: a5 = 20; a8 = 4/25.

Is that the complete problem statement? If so, then I am as lost as you!

There has to be some rule in effect here, because there is no way to deduce the terms of a series from only two points. If you could state that rule, it would be helpful.

Sn= (a1-a1r^n)/1-r where r= common ratio and n is the number of terms in the series. One of the problems was: find the sum of the series: a1= 3/5, r=2, n-7. When I did this, it turned out to be a horrible number, when in fact the answer was supposed to be 47 and 5/8.

There's no way the answer is 47 5/8. There are only 7 terms, so I added them up with a calculator (erased it before writing it down, though ).

 

[KillaMarcilla]

Yo, d00d, my math skillz are fairly l33t, but I can't remember anything about geometric series, and can't find my big math textbook (all I could find was one that said that the infinite series x^k converges to 1/(1-x) if x < 1, and diverges otherwise .. not too useful)

But, if you can bust out with a more thorough description, I can probably help you out

 

[bogdan]

a8=q*a7=q^2*a6=q^3*a5;
q^3=a8/a5=(4/25)/20=1/125;
q=1/5...
Mentor Edit: Rest of solution removed.[/color]

a1=2;a2=5;q=a2/a1=5/2;
an=625/8=a1*q^(n-1)=2*(5/2)^(n-1);
Mentor Edit: Rest of solution removed.[/color]

a1= 3/5; r=2; n=7

Sn= a1+a2+...+an;
Sn=a1+a1*r+a1*r^2+...+a1*r^(n-1); |*r
r*Sn= a1*r+a1*r^2+...+a1*r^(n-1)+a1*r^n; |-

Sn*(r-1)=a1*(r^n-1);
Mentor Edit: Rest of solution removed.

If you insert the numbers in and calculate, you will get 76.2.[/color]
...

 

[quantumdude]

Bogdan, I appreciate that you want to help, but please do not post complete solutions here. Astronomer 107 has to learn this stuff for herself.

 

[bogdan]

Oops..sorry...that will never happen again...