Question 1006888
Each term is 1/3 the previous term.
{{{A[n]=2187/3^(n-1))}}}
So then,
{{{2187/3^(n-1)=1/9}}}
{{{3^(n-1)=19683}}}
{{{n-1=log(3,19683)}}}
{{{n-1=log((19683))/log((3))}}}
{{{n-1=9}}}
{{{n=10}}}
It's the tenth term.
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{{{A[n]=2187/3^(n-1)}}}
{{{A[n]=2187*(1/3)^(n-1)}}}
{{{A[n]=2187*(1/3)^(n)*(1/3)^(-1)}}}
{{{A[n]=6561*(1/3)^n}}}
So then the partials sums are,
{{{S[n]=6561*((1/3)(1-(1/3)^n))/((1-1/3))}}}
{{{S[n]=(6561/2)*(1-(1/3)^n)}}}
When {{{n=5}}},
{{{S[5]=(6561/2)(1-(1/3)^5)}}}
{{{S[5]=(6561/2)(1-1/243)}}}
{{{S[5]=(6561/2)(243/243-1/243)}}}
{{{S[5]=(6561/2)(242/243)}}}
{{{S[5]=3267}}}