Question 1082960
A geometric progression (called a geometric sequence in the USA) is a sequence of numbers (called terms) where the ratio of consecutive terms is always the same. That ratio is called the common ratio, {{{r}}} .
In this case {{{r=8/12=2/3}}} .
If we call the first term of our sequence {{{b}}} ,
the first {{{n}}} terms of the progression are
{{{b}}} , {{{b*r}}} , {{{b*r^2}}} , {{{b*r^3}}} , ..., {{{b
*r^(n-1)}}} .
The sum of those {{{n}}} terms is
{{{b*(1+r+r^2+r^3+" ..."+r^(n-1))=b*((r^n-1)/(r-1))}}} .
When {{{r<1}}} , {{{r-1<0}} and {{{^-1<0}}} , so we like to write it as
{{{b*((1-r^n)/(1-r))}}} instead.
In that case, as {{{n}}} increases, {{{r^n}}} becomes smaller,
tending to zero.
So the sum to infinity is {{{b*((1-0)/(1-r))=b/(1-r)}}} .
In this case, the third term is
{{{b*r^2=12}}} , or {{{b*(2/3)^2=12}}} , or {{{b*(4/9)=12}}} .
So, {{{b=12*9/4=27}}} , and the sum to infinity is
{{{b/(1-r)}}}={{{27/((1-2/3))}}}={{{27/((1/3))}}}={{{27*3=highlight(81)}}} .