Question 87238
a)
The ratio r is the factor to get from term to term. So to find r, simply pick any term and divide it by the previous term:

{{{r=(1/27)/(1/9)=(9/27)=1/3}}}  Divide the term of {{{1/27}}} by {{{1/9}}}

So the ratio is
{{{r=1/3}}}
The sequence is reduced by a factor of {{{1/3}}} each term, so the sequence is {{{(1/3)^n}}}


b)
The sum of a geometric series is
{{{S=a(1-r^n)/(1-r)}}}where a=1


{{{S=(1-(1/3)^10)/(1-(1/3))}}} So plug in {{{r=1/3}}} and  n=10 to find the sum of the first 10 partial sums


{{{S=(1-1/59049)/(1-(1/3))}}} Raise {{{1/3}}} to the 10 power


{{{S=(59049/59049-1/59049)/(1/2)}}} Make "1" into an equivalent fraction with a denominator of 59049


{{{S=(59048/59049)/(1-(1/3))}}} Subtract the fractions in the numerator


{{{S=(59048/59049)/(2/3)}}} Subtract the fractions in the denominator


{{{S=177144/118098}}} Flip the 2nd fraction, multiply, and simplify


So the sum of the first ten terms is {{{177144/118098}}} or 1.49997459736829 approximately


note: I chose to use fractions (to maintain accuracy), but it may be much easier for you to simply use a calculator to evaluate the sum.

c)
The sum of a geometric series is


{{{S=a(1-r^n)/(1-r)}}}where a=1


{{{S=(1-(1/3)^12)/(1-(1/3))}}} So plug in {{{r=1/3}}} and  n=12 to find the sum of the first 12 partial sums


{{{S=(1-1/531441)/(1-(1/3))}}} Raise {{{1/3}}} to the 10 power


{{{S=(531441/59049-1/59049)/(1/2)}}} Make "1" into an equivalent fraction with a denominator of 531441


{{{S=(531440/531441)/(1-(1/3))}}} Subtract the fractions in the numerator


{{{S=(531440/531441)/(2/3)}}} Subtract the fractions in the denominator


{{{S=1594320/1062882}}} Flip the 2nd fraction, multiply, and simplify


So the sum of the first twelve terms is {{{1594320/1062882}}} or 1.49999717748537 approximately


d)
It appears that the sums are approaching a finite number of {{{3/2}}} or 1.5. This is because each term is getting smaller and smaller. This observation is justified by the fact that if {{{abs(r)<1}}} then the infinite series will approach a finite number. In other words
If {{{abs(r)<1}}} (the magnitude of r has to be less than 1) then,
{{{S=a/(1-r)}}}Where S is the sum of the infinite series. So if we let a=1 and r=1/3 we get
{{{S=1/(1-(1/3))}}}
{{{S=1/(2/3)}}}
{{{S=3/2}}}So this verifies that our series approaches {{{3/2}}}.