Question 326512
The infinite sum is given by {{{a1/(1-r)=18}}}

The nth partial sum is given by {{{a1((1-r^n)/(1-r))}}}

Since we are told the first 2 terms sum to 10, then we have 

{{{a1((1-r^2)/(1-r))=10}}}

But, from the first equation, {{{a1=18(1-r)}}}

Sub into the second equation:

{{{18(1-r)((1-r^2)/(1-r))=10}}}

{{{r=2/3}}}

and the first term, a1, is

{{{18(1-2/3)=6}}}

Since the nth term is gven by {{{a1*r^(n-1)}}}, 

The fourth term is

{{{6*(2/3)^3=16/9}}}