How do you find #S_n# for the geometric series #a_1=2#, #a_6=486#, r=3?

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Answer 1 · 2017-05-10T10:40:48

Sum of the series for #6# terms is #728#

Number of term of geometric series is #n=6#

Common ratio of geometric series is #r=3#

1st term of geometric series is #a_1=2#

6th term of geometric series is #a_6= a_1* r^(n-1) =2*3^(6-1)=2*3^5= 486#

Sum of the series formula : #S_n = a_1* ((1-r^n)/(1-r))#

Sum of the series for #6# terms : #S_6 = 2* ((1-3^6)/(1-3)) = #

#cancel2* ((1-3^6)/ cancel(-2)) = - (1-3^6) = -(1-729) =728#[Ans]