anonymous
  • anonymous
Find the general term an for the geometric sequence a1=-3 and a2=6?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[a_{1}=-3 and a_{2}=6\]
RadEn
  • RadEn
use the formula : an = a1 r^(n-1) with a1 is given -3 and to get the ratio(r), defined r = a2/a1 = 6/-3 = -2 therefore, an = a1 * r^(n-1) an = -3 * (-2)^(n-1) |dw:1367598379663:dw|
anonymous
  • anonymous
so that's the answer? @RadEn I'm confused

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nincompoop
  • nincompoop
it is not the answer, rather hint as how to solve the problem.
anonymous
  • anonymous
ahhh!! i'm so lost and confused!! @nincompoop
anonymous
  • anonymous
Can you find here the common ratio "r" ??
anonymous
  • anonymous
-2?
anonymous
  • anonymous
Yep..
anonymous
  • anonymous
So, general term can be found as: \[\large \color{green}{a_n = a \cdot r^{n-1}}\]
anonymous
  • anonymous
a = -3 and r = -2.. Plug in and find..
anonymous
  • anonymous
6?
anonymous
  • anonymous
How?
anonymous
  • anonymous
Use that formula,
anonymous
  • anonymous
-3(-2)^n-1?
anonymous
  • anonymous
Yep, this is the general term you require...
anonymous
  • anonymous
okay, but I am stuck here. I dont what is the next step
anonymous
  • anonymous
i get 6^n-1?
anonymous
  • anonymous
Hey, this is the required answer, you can't solve it further..
anonymous
  • anonymous
so this cant get be simplified any more?
anonymous
  • anonymous
No..
anonymous
  • anonymous
\[\large a_n = -3\cdot (-2)^{n-1}\]
anonymous
  • anonymous
okay... how come tho? 6^n-1 looks like an ideal answer to me.
anonymous
  • anonymous
You cannot multiply \(-3\) with \((-2)^{n-1}\)
anonymous
  • anonymous
ohhhhhhhhhhh!!!!!!!!!! duh!! now i get it! my brain hurts been studing all week.
anonymous
  • anonymous
It is okay..
anonymous
  • anonymous
thank you for your time and patience
anonymous
  • anonymous
You are welcome dear..

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