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Answer 1 · 2018-02-26T22:30:13

#2/27#

to find #a_4#, set #n# to #4#.

#a_n = 6 * (1/3)^n#

#a_4 = 6 * (1/3)^4#

#(1/3)^4 = 1/(3^4)#

#= 1/81#

#6 * (1/3)^4 = 6 * 1/81#

#= 6/81#

#6/81# can be simplified to #2/27#.

Answer 2 · 2018-02-26T22:47:44

#a_4=2/27#

#"to evaluate "a_4" substitute n = 4 into the formula"#

#rArra_(color(red)(4))=6xx(1/3)^(color(red)(4))#

#color(white)(xxxx)=6xx1/81=6/81=2/27#