What is the 6th term of the geometric sequence where a1 = 128 and a3 = 8?
i just need a formula, is all
so, whats the problem? Use a formula?
yes, that's what i just asked for..
@nincompoop
oh, so you can google it, can't you? http://en.wikipedia.org/wiki/Geometric_progression#Elementary_properties
i didnt understand it which is why I'm here....seeking help
So, you know what formula is \[a ^{n}=a _{1}q ^{n-1}\]
@rajathsbhat @Kamille i don't get the formula is my problem
does this help ? http://www.regentsprep.org/Regents/math/algtrig/ATP2/GeoSeq.htm
the common ratio, how do you find that?
oh, it should be \[a _{n}=a _{1}q^{n-1}\] Well, to get the formula is very easy. Let's say, that there is a number a1 then a2=a1*q, a3=a2*q etc, do you understand why?
\[\Large a_{3}=a_{1}q^{2}\] right? you have a3 and a1. Substitute them into the equation to find q. You can then find a6.
geometrical progression means that very is a sequance, where a1;a2(a1*q);a3(a2*q), etc when q=a2/a1
=0.125
isn't it 0.25 or 1/4?
it is 1/4
well, why it can't be -1/4?
q must be >0?
yep,equally possible
no its 0.125
thanks guys
you have forgoten to square root the answer, yummydum
i put 0.125 and it was right...
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