anonymous
  • anonymous
What is the sum of the arithmetic sequence 3, 9, 15..., if there are 36 terms?
Mathematics
chestercat
  • chestercat
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welshfella
  • welshfella
Sum of n terms = (n/2)[2a + d(n-1)] where d = common difference and a = first term
welshfella
  • welshfella
common difference d = 9- 3 = 6
anonymous
  • anonymous
right then what do I do?

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welshfella
  • welshfella
well plug in the values n = 36, d = 6 and a = 3
welshfella
  • welshfella
plug these in and work it out
welshfella
  • welshfella
= (36/2)[2*3 - 6(36-1)]
welshfella
  • welshfella
sorry that should be a + before the 6 not -
anonymous
  • anonymous
you're all good so then I have to solve that?
welshfella
  • welshfella
do the parts in the parentheses first
anonymous
  • anonymous
(18)[35] is what I got but that doesn't seem like it makes sense at all
welshfella
  • welshfella
36/2 = 18 36 - 1 = 35
welshfella
  • welshfella
no thats not right its 18(6 + 6*35) do the stuff in the parentheses first then finall y multiply by 18
anonymous
  • anonymous
7560?
welshfella
  • welshfella
multiply 6 by 35 before adding the 6
anonymous
  • anonymous
actually 3888
welshfella
  • welshfella
no i think you are doing things in the wrong order remember PEDMAS Parentheses Exponent Division Multiplication Add Subtract thats the order of operations
welshfella
  • welshfella
yes 3888 is correct 18(6 + 6*35 = 18*(6 + 210) = 18 * 216 = 3888
anonymous
  • anonymous
thank you!
welshfella
  • welshfella
PEDMAS is very important
welshfella
  • welshfella
yw
welshfella
  • welshfella
note i worked out 6 + 210 before i multipilied by 18 because 6 + 210 was in the parentheses

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