Results are out! Find what you Get quick advice or join the chat
Hey! Sign in to get help with your study questionsNew here? Join for free to post

c2 geometric series

Announcements Posted on
Talking about ISA/EMPA specifics is against our guidelines - read more here 05-03-2015
  1. Offline

    a geometric progression has a positive common ratio, its first terms are 32,b,12.5 how do you find the value of b
    could you please show step by step instructions so i know what technique to use when answering similar questions
  2. Online

    If you know two consecutive terms of a gp you can find the value of r by dividing
  3. Offline

    (Original post by the bear)
    If you know two consecutive terms of a gp you can find the value of r by dividing
    they arent consecutive
  4. Online

    Sorry i misread 12.5 as 12, 5...

    Instead say

    b/32 = 12.5/b..... both sides are the ratio r

    solve for b
  5. Offline

    (Original post by anisah4295)
    they arent consecutive
    From your post, it implies they are. Is the first term 32, with the second being b and the third being 12.5? Consecutive means 'in order', so 1, 2, 3 or similar.
  6. Offline

    Is it necessary to know how to prove the formula of sum to infinity in C2?
  7. Offline

    (Original post by Buongiorno)
    Is it necessary to know how to prove the formula of sum to infinity in C2?
    no, you only need to know the formula
  8. Offline

    write the series down and under the first term write A, AR under the second, AR^2 under the third and so on. you have the first and third terms so that A and AR^2, so divie the third by the first and you're left R^2. square root and you have the common ratio
  9. Offline

    But why use a formula if you don't know how it works? Waste of math in my opinion! And the formula is dead simple!

    Ok, so take the sum of n terms of a sequence starting with a and common factor r.

    Sn = a(r^n -1)/(r-1)

    Multiply top and bottom by (-1), Sn = a(1 - r^n)/(1 - r)

    Now think about it... what happens when n goes to infinity? Well... if r>1, then r^n gets bigger and bigger to infinity. Same thing when r<(-1) (but negative), so Sn goes to infinity!

    Now what happens if r=1? It becomes a(1-1)/(1-1). But wait... we can't divide by 0! So this doesn't work.
    With r=(-1), this gets really weird, because when n is even, it becomes a(1-1)/(1- (-1)) = 0.
    But when n is odd, this becomes, a(1-(-1))/(1-(-1)) = a. But which is infinity? Even, or odd? So we don't really know. This is undefined.

    What if -1<r<1 ? Well... then r>r^2>r^3>....r^n>...., so when n goes to infinity, r^n becomes 0! :O If we plug that in, we get Sn = a(1-0)/(1-r) = a/(1-r)

    Therefore, when |r|<1, Sn = a/(1-r).

    See the truth is, if you don't understand what you're doing in math, you WILL get a LOWER grade, and you WILL have trouble. For example, I don't have to memorize formulas like that, because I know how to prove it! So if I REALLY get stuck, I can just quickly prove it, half in my head, half on paper, in like 2 mins. And then you'll use the formulas that you got many times and eventually... POOF! Its hammered in your subconscious. And this takes a LOT longer to happen, if you just try to learn the formula.


Submit reply


Thanks for posting! You just need to create an account in order to submit the post
  1. this can't be left blank
    that username has been taken, please choose another Forgotten your password?
  2. this can't be left blank
    this email is already registered. Forgotten your password?
  3. this can't be left blank

    6 characters or longer with both numbers and letters is safer

  4. this can't be left empty
    your full birthday is required
  1. By joining you agree to our Ts and Cs, privacy policy and site rules

  2. Slide to join now Processing…

Updated: April 27, 2012
2015 general election
New on TSR

Loved by Students

Our big survey results unveiled

Article updates
Quick reply
Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.