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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.

Liam
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1 Answers1

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

Quality
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  • Apparently the correct answer for the sum of the first 7 elements is: 4*z^7/(z-1)-4/(z-1). Not quite sure why though. Thanks – Liam Mar 12 '15 at 02:29
  • Hm, il take a look at different ways you can write it and edit it – Quality Mar 12 '15 at 02:45
  • It can become confusing depending on notation as $a_0$ is usually the first term. How many terms are they wanting in the sum? – Quality Mar 12 '15 at 02:47