How do you find #a_3# for the geometric series given #S_n=249.92#, r=0.2, n=5?

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Answer 1 · 2017-11-21T12:23:18

# a_3=8#

#S_n=249.92 , r =0.2 , n=5 #

#S_n= a_1 *(1-r^n)/(1-r) #

#:. 249.92= a_1 *(1-0.2^5)/(1-0.2) #

#:. a_1 = {249.92*(1-0.2)}/ (1-0.2^5)=200#

First term is #a_1=200#

Third term is #a_3=a_1 * r ^(n-1)= 200 * (0.2)^(3-1)# or

#a_3== 200 * (0.2)^2=200* 0.04=8#

# :. a_3=8# [Ans]