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Given the sequence –3, 0, 3, which of the following is equivalent to:

Mathematics
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\[\sum_{n=5}^{8}A -subscript- N\]
54 102 144 162

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Other answers:

First term is -3, common difference is 0-(-3) = 3 So, the sequence is \[a_n = -3+3(n-1)\]
so, we can do that plugging n-2, n-3 etc. into the next equations?
[-3+3(n-1)]+[-3+3(n-2)] etc.?
You need to find sum from the fifth term to the eighth term, so, find a5, a6, a7, a8 and add them together.
n is the number of term
that's right we start on 5 right? so like this?
e.g. \[a_5 = -3+3(5-1)=...\]
Start from 5, yes.
[-3+3(n-5)]+[-3+3(n-6)]+[-3+3(n-7)]+[-3+3(n-8)]?
(-3+3n-15)+(-3+3n-18)+ (-3+3n-21)+ (-3+3n-24)?
am I doing this right?
(3n-18)+(3n-21)+(3n-24)+(3n-x)+(3n-27)
i'm not sure what to do from there
No
\[a_5 = -3+3(5-1)\]\[a_6 = -3+3(6-1)\]\[a_7 = -3+3(7-1)\]... Then, find \(a_5+a_6+a_7+a_8\)
=−3+3(5−1) =−3+3(6−1) =−3+3(7−1) =-3+3(8-1)
=-6+15 =-6+18 =-6+21 =-6+24 =9 =12 =15 =18 9+12+15+18
54
thank you so much guys

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