someone explain to me!
posted by Andrew .
the third term of a geometric sequence is t3= 75 and the sixth term is t6=9375. determine the first term and the common ratio.
how do I do this question I am tying really hard but I still don't comprehend this question!
We have t_{3} = 75 and t_{6} = 9375.
A geometric sequence looks like
a, ar ar_{2}, ar_{3}, ar_{4}, ar_{5}, ar_{6}...
If you take the ratio of ar^{5} to ar^{2} you should get t_{6}/t_{3}. Note that the exponent of the ratio is always one less than the term number. In any case, this is ar^{5}/ar^{2} = r^{3}
so look at 9375/75 = r^{3} = something you can do
Note that this is an alternating sequence.
Can you take it from here?
I don't know about andrew but I am looking at a similar question like this and was wondering if you could show how to do the rest thanks!
I'll try to get you started, or help if you're stuck somewhere, butI won't do the rest. :[
This right here
a, ar, ar_{2</subp>, ar3</subp>, ar4</subp>, ar5</subp>, ar6</subp>
should've been
a, ar, ar2, ar3, ar4, ar5, ar6
These are the first 7 terms of the sequence. The first term is always ar0 = a
I do a lot of copy/paste and sometimes don't proofread what I'm writing until after it's posted.
Anyway, I hope you can follow what I outlined.
I understand! thanks!
hmmm...looks like I have too many distractions right now...I'm forgeting to check that the tags are balanced.
This here
This right here
a, ar, ar2, ar3, ar4, ar5, ar6
should be
I'm glad you made sense of the post in spite of typos.
a, ar, ar2, ar3, ar4, ar5, ar6
}
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