anonymous
  • anonymous
what are the explicit equation and domain for a geometric sequence with a first term of 5 and a second term of -10
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
anonymous
  • anonymous
jim_thompson5910
  • jim_thompson5910
5r = -10 r = ???

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anonymous
  • anonymous
-2
jim_thompson5910
  • jim_thompson5910
First Term: a = 5 Common Ratio: r = -2 Plug those into \[\Large a_{n} = a*(r)^{n-1}\] to get the general nth term formula
anonymous
  • anonymous
5n = 5 x (-2) n-1 5n = -10 n-1 What do i do from here
myininaya
  • myininaya
no a_n symbolizes the n term of the sequence a_n and a are different values also we don't have (ar)^(n-1) we just have that the r is to the (n-1)
anonymous
  • anonymous
Ok, so how do i set up this equation
myininaya
  • myininaya
\[a_n=a \cdot r^{n-1}\] just replace the a with 5 and replace the r with -2 like jim said
anonymous
  • anonymous
5n = -10^n-1 ?
myininaya
  • myininaya
again a_n and a are different values a_n is a variable by itself you can't multiply a and r because r has an exponent
anonymous
  • anonymous
5_n = 5 x -2^n-1
myininaya
  • myininaya
you cannot replace a_n with 5_n this makes no sense let me show you how this formula is come up with so maybe you understand better {a_n} is a sequence of numbers those numbers are a_1=5 a_2=5(-2)=-10 a_3=5(-2)(-2)=5(-2)^2=20 a_4=5(-2)(-2)(-2)=5(-2)^3=-40 a_5=5(-2)(-2)(-2)(-2)=5(-2)^4=80 ... therefore a_n=5(-2)(-2)(-2)(-2)(-2)(-2)...(-2)= (by the way that is a (n-1) amount of (-2)) I put a (n-1) amount of (-2) because looking at a_1 we have (-2)^0=(-2)^(1-1) looking at a_2 we have (-2)^1=(-2)^(2-1) looking at a_3 we have (-2)^2=(-2)^(3-1) it is always one less than where the number is in the sequence
anonymous
  • anonymous
So how do i set up the equation ? im confused
anonymous
  • anonymous
Does the 5 not go where the n is
jim_thompson5910
  • jim_thompson5910
5 goes where 'a' is but \(\Large a_n\) is NOT the same as just 'a'
jim_thompson5910
  • jim_thompson5910
If it confuses you, think of it as \[\Large T = a*(r)^{n-1}\] T = nth term a = first term r = common ratio n = positive whole number (used to identify which term you're dealing with) example: n = 3 ---> 3rd term
anonymous
  • anonymous
Ok
jim_thompson5910
  • jim_thompson5910
The notation \(\Large a_n\) is used for sequences because the 'n' changes to a positive whole number to indicate a certain term example: 17th term ---> n = 17 ---> \(\Large a_n = a_{17}\)
anonymous
  • anonymous
Ok, so you leave the an as it is
myininaya
  • myininaya
Yep it is a formula for the nth term in the sequence
anonymous
  • anonymous
So it is an = 5 (-15)^n-1; all integers where n less than or equal to 1
myininaya
  • myininaya
how do you get -15 for r?
myininaya
  • myininaya
r symbolizes the geometric ratio to find r all you have to do is evaluate a_2/a_1 or take any two consecutive numbers in the sequence and put the 2 nd term of those numbers over the 1st term of those numbers
myininaya
  • myininaya
also this was already determined way above when joe was working with you
anonymous
  • anonymous
the answer is A!!!!!!!!!!!!!!!!!

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