Answers

2014-10-13T14:31:28+08:00
For geometric progression you'll have the formula:
a_n = a_1 r ^{n-1}
where:
        a1 is the first term = 3
        an is the last term which in this case is a6=32/81
        r is the common ratio
substituting the formula:
a_n = a_1  r^{n-1}
 \frac{32}{81} = 3 r ^{6-1}
 \frac{32}{81(3)} =r^5
extracting the fifth root you'll have
r =  \frac{2}{3}
a_2 = 3( \frac{2}{3} )
a_2 = 2
a_3 = 2( \frac{2}{3})
a_3 =  \frac{4}{3}
a_4 =  \frac{4}{3} (\frac{2}{3})
a_4 = \frac{8}{9}
a_5 = \frac{8}{9} ( \frac{2}{3} )
a_5 =  \frac{16}{27}
therefore you'll have the geometric series as:
3, 2, 4/3, 8/9, 16/27, 32/81
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