How do you find $a_6$ for the geometric sequence $540, 90, 15,...$ ?

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Answer 1 · 2016-10-23T10:27:37

The sixth term $a_6$ is $5/72$

In a geometric series, whose first term is $a_1$ and common ratio is $r$ , $n^(th)$ term $a_n$ is given by $a_1xxr^(n-1)$ .

In the given series ${540,90,15,....}$ , first term is $540$ and common ration $r$ is $90/540=15/90=1/6$

Hence the sixth term $a_6$ is $a_1xxr^(n-1)=540xx1/6^(6-1)$

= $540/(6xx6xx6xx6xx6)=(6xx6xx3xx5)/(6xx6xx6xx6xx6)$

= $(cancel6xxcancel6xxcancel3xx5)/(cancel6xxcancel6xx2cancel6xx6xx6)$

= $5/72$

How do you find a_6a_6 for the geometric sequence 540, 90, 15,...540, 90, 15,...?