If #a_n=1/3*6^(n-1)#, what is #a_6#?

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Answer 1 · 2016-10-20T09:47:13

#a_6=2592#

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

we can find #a_6# by substituting #n# by #6#.

Hence #a_6=1/3*6^(6-1)#

= #1/3*6^5#

= #(6xx6xx6xx6xx6)/3#

= #(2cancel6xx6xx6xx6xx6)/(1cancel3)#

= #2592#