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find the sum find the third term of the geometric sequence in which a1=36 and a6=4/27

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Kenshinbarredoid
in finding the 3rd term:
an=a1rn-1
4/27=36r6-1
4/27/36=36r5/36
4/972=r5
1/243=r5
5√1raised to 5/3 raised to 5=5√r5
1/3=r

an=a1rn-1
a3=36(1/3)3-1
a3=36(1/3)2
a3=36(1/9)
a3=4

sum of series

sn=a1-a1rn/1-r
s6= 36-36(1/3)6÷1-1/3
s6=36-36(1/729)÷2/3
s6=36-4/81÷2/3
s6=2912/81÷2/3
s6=4368/81

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