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Geometric Sequences and Series

  • #1
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!
 

Answers and Replies

  • #2
Tom Mattson
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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 :frown:).
 
  • #3
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
 
  • #4
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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.
I'm not sure why you set it up that way but I just worked that one out (and got an answer that works) so I'll just show you how I did it.

The formula for the general rule goes: an=a1rn-1 (so for the first term r is to the zeroth power).

To write down the general rule, you'll need to figure out the first term a1 and the common ratio r. You've been given two pieces of information so you can set up a system of two equations and figure out both variables.

You know that the fifth term (meaning n is 5) is 20 (meaning an is 20) so plug what you know into the general rule formula:

20 = a1r4

You also know that the eight term (n is 8) is 4/25 (an is 4/25). So plug that information in:

4/25 = a1r7

Now, can you solve that system (you'll have to use substitution)?


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.
Forget about that part I put in boldface for a second. Use the two blue pieces of information the same way I did above to find the first term (a1) and the common ratio (r) to formulate your general rule.

Once you get that, fill in the piece of boldfaced information and I believe you'll have to use logs to solve for n. Have fun. :wink:

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!
Are you sure about that answer? I get 76.2 but that doesn't feel right.
 
Last edited:
  • #5
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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.

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.

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.

...
 
Last edited by a moderator:
  • #6
Tom Mattson
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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.
 
  • #7
191
0
Oops..sorry...that will never happen again...
 

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