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Express the infinite series $$\frac{3}{206}+\frac{9}{2\cdot103^2}+\frac{27}{2\cdot103^3}+\cdots$$as a terminating decimal.

 

 

 

 

A particular geometric sequence has strictly decreasing terms. After the first term, each successive term is calculated by multiplying the previous term by $\frac{m}{7}$. If the first term of the sequence is positive, how many possible integer values are there for $m$?
 May 21, 2018
 #1
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sorry, can't read your question!.

 May 21, 2018
 #2
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Express the infinite series

 

\(\frac{3}{206}+\frac{9}{2\cdot103^2}+\frac{27}{2\cdot103^3}+\cdots\)

 

as a terminating decimal.

 

This is a ordinary GP

 

\(S_\infty=\frac{a}{1-r}\)

 

If you have done GPs this will be easy. If not then you better say so.

 May 21, 2018
 #3
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I did not

Guest May 22, 2018
 #4
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S =[3/206] / [1 - 3/103]

S =[3/206] x [103/100]

S = 309 /20,600

S = 3 / 200

S = 0.015

 May 22, 2018

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