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Mathematics 7 Online
OpenStudy (anonymous):

Find the next three terms of the sequence. Then write a rule for the sequence. 648, 216, 72, 24

OpenStudy (king):

8,4,2..... or 8,8/3,8/9.....

OpenStudy (king):

if u tell me which one is rite i shall tell u the rule.......

OpenStudy (king):

Rohan think and reply.....

OpenStudy (anonymous):

8,4,2..... or 8,8/3,8/9.... it will be going on on and onnnnnnnn

OpenStudy (anonymous):

u dont be a barrier

OpenStudy (anonymous):

the rule plzzzzzz!

OpenStudy (king):

which one is rite 8,4,2 or 8,8/3,8/9

OpenStudy (anonymous):

the second

OpenStudy (anonymous):

the first

OpenStudy (king):

@Rohangrr ure wrong once again so thats why i said think and reply!

OpenStudy (anonymous):

is the answer the 2nd

OpenStudy (king):

@daja2fly the rule is that see, 648 can be written as 2^3*3^4 216 can be written as 2^3*3^3 72 can be written as 2^3*3^2 24 can be written as 2^3*3^1 the next term will be 2^3*3^last power -1 then 2^3*3^0 then 2^3*3^-1 then 2^3*3^-2... and so on....!!!!!

OpenStudy (king):

understood?

OpenStudy (anonymous):

i get the sequence but not the actual rule how would i write that

OpenStudy (king):

u can write it as 2^3*3^last power-1 \[2^{3} \times 3^{x _{l}-1}\]

OpenStudy (king):

where \[x _{l}\] is the previous power

OpenStudy (king):

ok?

OpenStudy (radar):

This is a geometric series where the first term is 648 and the common ratio is 1/3. The next term would be 24(1/3) or 8, and the next is 8/3 etc. The expression for the nth term would be:\[a _{n}=(648)(1/3)^{n-1}\]. Substituting and solving for the fifth term:\[a _{5}=648(1/3)^{4}=648(1/81)=8\]

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