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# c2 geometric series

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Why bother with a post grad course - waste of time? 17-10-2016
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...

Instead say

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.

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