Question 942795
to find each term, we divide by {{{3}}}, a common ratio, from the previous term

{{{243/3=81}}}
{{{81/3=27}}}

so, next would be
{{{27/3=9}}}
{{{9/3=3}}}
{{{3/3=1}}}
{{{1/3=1/3}}} and so on

we can model it {{{243}}}, {{{81}}}, {{{27}}},{{{ 9}}}, {{{3}}},{{{1}}}... with the exponential function below:

{{{a[n]=3^(-n+ 6) }}} where {{{n}}}={{{1}}},{{{ 2}}}, {{{3}}},{{{4}}}...

Check:

When {{{n = 1}}}, which represents the first term, we get 
{{{a[1]= 3^(-1+ 6) = 3^5 = 243}}}

When {{{n = 2}}}, which represents the second term, we get 

{{{a[2]=3^(-2+ 6) = 3^4 = 81}}}