Find the 17th term of the sequence a_18 = 32, d=-4
18a = 32 32 / 18 = a a = ?
1.78
That solves for a. I've no clue what you mean by 17th term, as it's not a sequence, and I don't see the point to value d.
I'm not sure either, that's what the problem in the textbook is asking
It says in the back of the book that it is 36
Quite a poorly worded question.
I had a problem earlier asking for the 32nd term for the sequence, but it actually gave me a list of numbers as this one doesn't
general form for arithmetic sequence: \[a_{n}=a_{1}+(n-1)d\] since we know 18th term we can substitute: \[(32)=a_{1}+((18)-1)(-4)\] solve for a1
Uhmm...I got -2.125 ?
Did I do it wrong?
\[32 = a_{1} - 68\] \[a_{1} = 100\]
Ohh, I wasn't paying attention and divided them instead of adding
@brusmack what do I do now?
now you can fill in the general equation \[a_{n}=100+(n-1)(-4)\]
use n=17 for the 17th term
Okay, so would it then be like \[a_{17}=100+(18-1)(-4)\]
Wait 17 not 18
yep, so you should get a(17)=36
Okay, thank you so much!!
no problem
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