PLEASE HELP I have tried to work it out but just cant... What is the 32nd term of the arithmetic sequence where a1 = 13 and a13 = −59? −185 −179 −173 −167
The nth term in general is an = a1 + d(n-1) where d is the common difference
i think 6 is the common difference...
a1 = 13 so, an = a1 + d(n-1) an = 13 + d(n-1)
-6 i mean
then we also know that a13 = -59 (ie the 13th term is -59) an = 13 + d(n-1) a13 = 13 + d(13-1) ... replace n with 13 -59 = 13 + d(13 - 1) solving that for d gives you d = -6, correct
and then that is where i got stuck at...
so that means an = a1 + d(n-1) an = 13 + d(n-1) an = 13 + (-6)(n-1) an = 13 - 6n + 6 an = -6n + 19
an = -6n + 19 is the nth term formula that allows you to find any term you want the 32nd term, so plug in n = 32 and evaluate
so a32+-6(32)+19=... right?
a32 = -6(32) + 19 I think you meant
yeah, my bad, i was pushing the shift key....
that's ok
now just evaluate -6(32) + 19
-211 is what i got... i did something wrong.
You should have gotten -173
alternatively, you can plug n = 32 into an = 13 + (-6)(n-1) to get an = 13 + (-6)(n-1) a32 = 13 + (-6)(32-1) a32 = -173
hmm you must have done 6*32 + 19 and made the result negative
i got lost at the part where its a32=13+(-6)(32-1)... and yeah, i thought thats what i was supposed to do because () usually indicates *...
right? so what was i actually supposed to do for this equation? @jim_thompson5910
either equation works since an = 13 + (-6)(n-1) is equivalent to an = -6n + 19
ohh, okay that makes much more since now!!
so instead of using () for multiplication, i use it for addition?
no it means multiplication (-6)(n-1) means -6 times n-1
you may need to use the symbol * for multiplication
BEHHHH I HATE MATH~
yeah it's picky sometimes
yes! VERY! i think i may understand it...
that's great, glad you do
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