Mathematics
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OpenStudy (lukecrayonz):
Find the nth term of the geometric sequence:
a_2 or a[2]=3, a_5 or a[5]=(3/64), n=1
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OpenStudy (lukecrayonz):
@satellite73
OpenStudy (lukecrayonz):
@bahrom7893 anything? haha
OpenStudy (lukecrayonz):
I literally know the equation if you have a[1], but nothing when you're solving for a[1]
OpenStudy (lukecrayonz):
Because the equation while using a[1], is a[n]=a[1]r^(n-1)
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OpenStudy (bahrom7893):
a2=a1*r=3; a1=3/r
OpenStudy (bahrom7893):
just do the same thing as the solved example.. or is there a trick somewhere?
OpenStudy (lukecrayonz):
And solving for r now.. other example doesn't make that much sense to me, and I feel like there is a trick because n-1, 1-1=0.
OpenStudy (lukecrayonz):
and anything to the power of 0 is 1, so a[1]*1=a[1]..?
OpenStudy (bahrom7893):
r is either powers of 4 or 2..
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OpenStudy (lukecrayonz):
Hmm..?
OpenStudy (bahrom7893):
hang on..
OpenStudy (bahrom7893):
r=1/4, a1=3/1/4=12
OpenStudy (bahrom7893):
a2=3=a1r
a1=3/r
OpenStudy (bahrom7893):
a5=a1r^4=a1rr^3=3r^3=3/64
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OpenStudy (bahrom7893):
r^3=1/64
r=1/4
OpenStudy (bahrom7893):
a1=3/r=3/1/4=12
OpenStudy (lukecrayonz):
Got it! Thank you so much haha, stumped me
OpenStudy (bahrom7893):
i don't even know why they give u n, it's totally irrelevant imo
OpenStudy (lukecrayonz):
Got another question, that literally has nothing to do with anything..
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OpenStudy (lukecrayonz):
Posted it