sequences and series - Consider the geometric progression ...

I think I'm on the right track with this but not entirely confident.

$a_1 = 4$ , $a_2 = 4z$ , $a_3 = 4z^2$, ...

1. The 6th element $a_6$


2. The sum of the first 7 elements

I'm sure this works differently to arithmetic progression and uses ratios but a little stuck even with Googling like mad.

Thanks in advance.

Answer

Okay, so we need to find out what there is in common( common ratio). You do this by dividing. notice $$4z/4=z, 4z^2/4z=z $$ so it is clear that our formula is $a_n=4z^{n-1}$

So what is $a_6$?

$a_6=4z^{6-1}=4z^{5}$

To sum, we use the formula $ \sum_{k=0}^{n-1}(ar^k)=$a($\frac{1-r^n}{1-r})$

Where a is the first term in the series, and r is common ration. n is what you want to sum up to.

Specifically, for the first 7 elements, we have n=7,

so $\sum_{k=0}^{6}(4z^k)$=$$4(\frac{1-z^7}{1-z})$$