Question 1194957

{{{9}}}, {{{3}}}, {{{1}}}, {{{1/3}}},...

The formula for nth term of a geometric progression is


{{{a[n]=a[1]*r^(n-1)}}}

given:

{{{a[1]=9}}}

{{{r=3/9=1/3}}}

{{{a[n]=9*(1/3)^(n-1)}}}  where {{{n}}}={{{1}}},{{{2}}},{{{3}}},....