anonymous
  • anonymous
Find an equation for the nth term of the arithmetic sequence. -15, -6, 3, 12, ...
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
Find a pattern. What are you doing to each number to produce the next in the sequence?
anonymous
  • anonymous
plus 9
anonymous
  • anonymous
-15 + 9(n - 1)

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More answers

DanJS
  • DanJS
yeah, that type has constant change through one term to next
anonymous
  • anonymous
so i'm right?
DanJS
  • DanJS
yeah you can test it, put in values of n, does it generate that sequence
anonymous
  • anonymous
thank you!! can u help with another one
anonymous
  • anonymous
Find an equation for the nth term of the arithmetic sequence. a14 = -33, a15 = 9
DanJS
  • DanJS
sure
DanJS
  • DanJS
that type of sequence, like i said has a cponstant change from one term to the next. like the +9 in the last one..
anonymous
  • anonymous
\[a _{14}=-33 \] and \[a _{15} = 9\]
DanJS
  • DanJS
here they gave you consecutive terms, and told you the sequence is arithmetic
DanJS
  • DanJS
the constant change from one term to the next is 42
anonymous
  • anonymous
yes
anonymous
  • anonymous
so \[a _{n}\]= -579 + 42(n - 1) ??
anonymous
  • anonymous
or is it an = -579 + 42(n + 1)
DanJS
  • DanJS
f the initial term is a1 and the common difference is d, then the nth term of the sequence is given by; an = a1 + (n-1)d - - - By taking the above result further, the nth term can be given also as; an = am + (n-m)d, where am is a random term in the sequence such that n > m.
anonymous
  • anonymous
is my answer right then
DanJS
  • DanJS
an = am + (n-m)*d an = -33 + (n - 15) *42
anonymous
  • anonymous
that isnt one of my answer choices
DanJS
  • DanJS
that more general form for any 2 terms, not just 1 and 2
DanJS
  • DanJS
yeah fix it up , an = initial term a0 + (n -1)*d
DanJS
  • DanJS
an = -33 + (n-14)*42 an = -33 + (n-1-13)*42 an = -33 + (n-1)*42 - (13*42) an = -579 + (n-1)*42
DanJS
  • DanJS
the initial term a0 is -579
anonymous
  • anonymous
thx
DanJS
  • DanJS
welcome

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