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MATH HELP!!!
GeT_RiGhT | 
Canada GeT_RiGhT_A_EIZO 
If 5 is the 1st term and 2560 is the 10th term, in a geometric sequence. What is the 15th term?
2016-12-20 20:21
godsent
2016-12-20 20:24
#4
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Iceland fatboislim 
king Most clutches 9
2016-12-20 20:28
#5
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Poland Drunkenstein 
an=a1*q^(n-1) find q and use it again ezaf
2016-12-20 20:29
#6
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Indonesia holyjoe 
5*x^9=2560 x^9=512 x=2 5*x^14=81920
2016-12-20 20:33
n1
2016-12-20 20:36
#7
NEO | 
Poland Trausky 
A geometric progression has general form ar,ar^2,…ar^(n-1), where a is the first term and r the common ratio. The sum to n-terms is a(1-r^n)/(1-r)
2016-12-20 20:36
how do you get the ratio?
2016-12-20 20:45
#26
NEO | 
Poland Trausky 
a=5 for 10th term: 5r^9=2560 r^9=2560/5 r^9=512 r= 512^(1/9) r=2 now for 15 th term 5r^14=x x=5 * 16384 x=81920
2016-12-20 21:34
an = a1 * q^(n-1) to get the quotient: 10th element: first element times the quotient on the 9th power means 2560 = 5*q^9 from this: q=2 15th element = 5*2^14 = 81920
2016-12-20 20:35
Why log 512 / log 9 give me 2.84 not 2??? I thought this is how you find the exponent
2016-12-20 20:49
why would you use logarithm we need 'q' from: " 512=q^9 " so the logarithm would look like: log(smallindex'x')(upperindex'512') = 9 and thats not really helping. just "9root" the 512 and you get 2, simple as that
2016-12-20 20:56
because natural log
2016-12-20 21:14
#9
Brehze | 
CIS lepestos 
An=A1*q^(n-1), where u can find q as (10-1)=9th root of 2560/5 (actually, q=2), A1=5, n -number of the term(15)
2016-12-20 20:36
op, in what kind of NA university do you learn this complicated stuff
2016-12-20 20:37
grade 11 math
2016-12-20 20:44
Dude i dont even know what geometric sequence is
2016-12-20 20:54
3rd world education, failing PISA test, orbán to the max at least no refugees
2016-12-20 20:56
1+1=2. /discuss
2016-12-20 20:57
#21
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Indonesia holyjoe 
1+1=10 fuck off stupid meat monkey
2016-12-20 21:11
1+1+1 = 11 jolyhoe
2016-12-20 21:15
666
2016-12-20 21:21
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