Question 957592
given:

the first {{{3}}} terms of a geometric progression are 
{{{a-1}}}, 
{{{a +3}}}, and 
{{{3a +1}}} 
for some positive number {{{a}}}

In a Geometric Sequence each term is found by {{{multiplying}}} the previous term by a {{{constant}}}. 

Let a {{{constant}}} be  {{{k}}}:

{{{(a -1)k= a +3}}}

{{{k= (a +3)/(a -1)}}}..........(1)

{{{(a +3)k= 3a +1}}}

{{{k= (3a +1)/(a +3)}}}..........(2)

make {{{(1)=(2)}}}

{{{(a +3)/(a -1)=(3a +1)/(a +3)}}} ............cross multiply

{{{(a +3)(a +3)=(3a +1)(a -1)}}}

{{{a^2 +3a +3a+9=3a^2-3a +a -1}}}

{{{a^2 +6a+9=3a^2-2a -1}}}

{{{0=3a^2-2a -1-a^2 -6a-9}}}

{{{2a^2-8a -10 =0}}}........simplify

{{{a^2-4a -5=0}}}

{{{a^2+a-5a -5=0}}}

{{{(a^2+a)-(5a +5)=0}}}

{{{a(a+1)-5(a +1)=0}}}

{{{(a-5)(a +1)=0}}}

solutions:

{{{a=5}}}

or 

{{{a=-1}}}

since given that first 3 terms are {{{a -1}}}, {{{a +3}}}, and {{{3a +1}}} for some {{{positive}}}
number {{{a}}}, our solution is {{{a=5}}}

so, your terms are:
{{{a -1=5-1=4}}}, 
{{{a +3=5+8=8}}}, and 
{{{3a +1=3*5+1=16}}}

now find {{{k}}}


{{{k= (a +3)/(a -1)}}}..........(1)
{{{k= 8/4}}}
{{{k= 2}}}

so, the fourth term is {{{2*16=32}}}