You are Here: Home

# c2 geometric series

Announcements Posted on
Fancy a fiver? Fill in our quick survey and we’ll send you a £5 Amazon voucher 03-05-2016
Talking about ISA/EMPA specifics is against our guidelines - read more here 28-04-2016
1. 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. If you know two consecutive terms of a gp you can find the value of r by dividing
3. (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. Sorry i misread 12.5 as 12, 5...

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

solve for b
5. (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. Is it necessary to know how to prove the formula of sum to infinity in C2?
7. (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. 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. 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.

## Register

Thanks for posting! You just need to create an account in order to submit the post
1. this can't be left blank
2. this can't be left blank
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
1. Oops, you need to agree to our Ts&Cs to register

Updated: April 26, 2012
TSR Support Team

We have a brilliant team of more than 60 Support Team members looking after discussions on The Student Room, helping to make it a fun, safe and useful place to hang out.

This forum is supported by:
Today on TSR

### iGCSE English Language

Here's the unofficial markscheme

Poll
Useful resources

### Maths Forum posting guidelines

Not sure where to post? Read here first

### How to use LaTex

Writing equations the easy way

### Study habits of A* students

Top tips from students who have already aced their exams