Asked by Ana....
which term of the sequence 18,12,8,.... is 512/729 is ?
Answers
Answered by
MathMate
1. find the common ratio:
r=T(n)/T(n-1)=12/18=2/3
2. form the general sequence
(considering 18 as first term, or n=1)
T(n)=ar^(n-1)=18(2/3)^(n-1)
3. solve for
T(n)=18(2/3)^(n-1)=512/729
or even simpler,
(2/3)^(n-1)=256/6561=(2/3)^8
Solve for n.
r=T(n)/T(n-1)=12/18=2/3
2. form the general sequence
(considering 18 as first term, or n=1)
T(n)=ar^(n-1)=18(2/3)^(n-1)
3. solve for
T(n)=18(2/3)^(n-1)=512/729
or even simpler,
(2/3)^(n-1)=256/6561=(2/3)^8
Solve for n.
Answered by
Steve
512/729 = 2^9/3^6
= 2^8/3^8 * 2*3^2
= 2^8/3^8 * 2*3^2
Answered by
Ana
Thanks :)
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