```Question 84120
a)
The ratio r is the factor to get from term to term. So
r=nth term/(n-1) term
{{{r=(1/8)/(1/4)=(4/8)=1/2}}}
{{{r=1/2}}}

So the ratio is {{{r=1/2}}}
The sequence is cut in half each term, so the sequence is {{{(1/2)^n}}}

b)
The sum of a geometric series is
{{{S=a(1-r^n)/(1-r)}}}where a=1
{{{S=(1-(1/2)^10)/(1-(1/2))}}}So plug in n=10 to find the sum of the first 10 partial sums
{{{S=(1-1/1024)/(1/2)}}}
{{{S=2046/1024}}}
So the sum of the first ten terms is {{{2046/1024}}} or 1.99805 approximately

c)
Use the same formula to find the sum of the 1st 12 terms
{{{S=a(1-r^n)/(1-r)}}}where a=1
{{{S=(1-(1/2)^12)/(1-(1/2))}}}So plug in n=12 to find the sum of the first 12 partial sums
{{{S=(1-1/4096)/(1/2)}}}
{{{S=8190/4096}}}
So the sum of the first twelve terms is {{{8190/4096}}} or 1.99951 approximately

d)
It appears that the sums are approaching a finite number of 2. 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 infinite series. So if we let a=1 and r=1/2 we get
{{{S=1/(1-(1/2))}}}
{{{S=1/(1/2)}}}
{{{S=2}}}So this verifies that our series approaches 2. ```