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PLEASE HELP URGENT ! what is the 7th of the geometric sequence where a1 = 3 and a5 = -768

Mathematics
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nothings THAT urgent :/
we should start by remembering the general formula for a geometric sequence
an = a1*r^(n-1)

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Other answers:

we know a1, and we know the value when n=5; from this we can grab the r value
can you fill those in for me?
kind of
give it a try, ill know if its right or not :)
the answer is going to be -3,072
it might be, but lets see if we can build up to that
ok
lets fill in what we are given: a1 = 3 and a5 = -768 when n=5 an = a1 * r^(n-1) a5 = -786 = 3 * r^(5-1) -786/3 = r^(4) -292 = r^(4) , 4rt the whole thing to find r 4rt(-292) = r as is this is not a possible sequence to me. an even root can be negative across the reals
an even root canNOT .... i passed typing class, honest :)
check your problem and make sure you havent got an error in the post
ok
i got a feeling the a1 = -3 for this to work like you want it to
yeah
it is
4rt(292) = r then this is what we get for r take the general formula again, knowing this r and n=7 a7 = -3 * (4rt(292))^(6) and a calculator should popout your intended results
can it be 12,288
no
...can what be?
a7 is going to be a negative value
-12,288
adjust for a mismath and i get: a7 = -3 * 256^(6/4) = -12288 , so yes
can you help me with thi one ? What is the sum of the geometric sequence -4, 28, -196 ?
why did you think that the answer "is going to be -3,072"?
i hd my ngatives wrng
signs
ah, itll prolly be best to post up a new question with that sum in it; i have to go for now

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