Determine whether the sequence converges or diverges. If it converges, give the limit. 60, -10, five divided by three, negative five divided by eighteen, ...
@jdoe0001
\(\bf \textit{sum of an infinite geometric serie}=\Sigma_{i=0}^\infty\quad a_1\cdot r^i=\cfrac{a_1}{1-r}\) so... .we know the 1st term term... is there.. just 60 to find "r" in a geometric sequence... just divide the "next term" by the "current term" and that will spit out the multiplier "r"
convergent... means it ends somewhere at some number divergent... means.. it doesn't have a limit, it just keep on going
so.. get "r" and plug it in at \(\large \bf \cfrac{a_1}{1-r}\) if it gives a value, that's it's limit, it ends there, thus is convergent
r=-6?
well.. not quite. close
\(\bf \cfrac{-\cancel{ 10 }}{\cancel{ 60 }}=?\)
6
is it positive 6?
-1/6?
hehe... yes.. -1/6 so there :)... just plug that in
\(\bf \cfrac{60}{1-\left(-\frac{1}{6}\right)}\)
360/7
yeap.... so it ends somewhere.. thus is convergent
wheres the limit?
360/7 <--
hmm, that's not of my answer choices :/
hmm 51.43 I'd think is the decimal of that
yes, but it is still not of my answer choices.
hmmm what are the choices given? I don't see anything wrong there
Diverges Converges; 11100 Converges; 72 Converges; 0
i think r was suppose to be positive, 1/6, so we get the limit 72. we are multiplying 1/6 everytime.
hmmm ....
so....... can't be.... is a negative 2nd and rth term... so is a negative "r"
2nd and 4th term I meant.... but.... hmmm nope.. looks to me ok with 360/7
maybe the negatives aren't meant to be there... or someone forgot to add the minus to the 1/6 when getting the choices done
i will ask my teacher about it.
it happens =)
everyone makes mistakes C;
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