anonymous
  • anonymous
In a geometric sequence, the term an+1 can be smaller than the term an. true or false?
Mathematics
chestercat
  • chestercat
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anonymous
  • anonymous
true
anonymous
  • anonymous
if n is < 1
anonymous
  • anonymous
are u sure?? how did u know so fast?? lol

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anonymous
  • anonymous
as in |n| < 1
anonymous
  • anonymous
so n ranges from 0 to 1
anonymous
  • anonymous
oops
anonymous
  • anonymous
wait
anonymous
  • anonymous
oh okay
anonymous
  • anonymous
if the interval is (-1) that makes it alternating
anonymous
  • anonymous
so it will be smaller and bigger alternatingly
anonymous
  • anonymous
(-1)^ 2 is positive, (-2) ^ 3 is negative
anonymous
  • anonymous
so this is FALSE?
anonymous
  • anonymous
(-2)^2 i mean
campbell_st
  • campbell_st
if the common ratio r <1 then each term is smaller that the previous
anonymous
  • anonymous
there are a few cases yeah. if the sequence convergerges or if the sign alternates
anonymous
  • anonymous
wait so its TRUE then???
anonymous
  • anonymous
yup its true
anonymous
  • anonymous
i said wait coz my explanation was incomplete
anonymous
  • anonymous
oh okay haha :P thanks!!! so its definitely TRUE right?
campbell_st
  • campbell_st
if the absolute value of r is less than 1 |r| < 1 the the geometric series will have a limiting sum...
campbell_st
  • campbell_st
just to clarify things... n is used for the term number... and r is used for the common ratio ( or multiplication constant)... just to avoid confusion.
anonymous
  • anonymous
k thanks for all the help y'all!
anonymous
  • anonymous
yeah. \[ar ^{n}\] if |r| < 1 it converges and if r is negative, it alternates
campbell_st
  • campbell_st
not quite a term in a geometric series is found using \[T_{n} = ar^{n -1}\] n is the term number, T is the Term a is the 1st term and r is the common ratio... please get it right @irkiz
anonymous
  • anonymous
k thanks for all the help y'all!

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