[anisah150]

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

[the bear]

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

[anisah150]

(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

[the bear]

Sorry i misread 12.5 as 12, 5...

Instead say

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

solve for b

[Contrad!ction.]

(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.

[Buongiorno]

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

[Stamaxo]

(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

[zack-w]

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

[bobbybuilder]

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.