Question 78942
a)
The ratio r is the factor needed to go from term to term. To find the factor, divide any term by its previous term. So I chose 1/3 as the first term to be divided by 1
Ratio r: {{{r=(nth term)/(n-1 term)}}}pick any nth term and any previous term, such as the 2nd and 1st term.
{{{r=(1/3)/(1)=1/3}}}I can also do it with 1/27 and 1/9 and it will still give me the same value
{{{r=(1/27)/(1/9)=(9/27)=(1/3)}}}
So r=1/3

b) The sum of the first ten terms can be found by using
{{{s=(1-r^(n+1))/(1-r)}}} 
So let r=1/3 
{{{s=(1-(1/3)^10)/(1-(1/3))}}}
{{{s=(1-(1/59049))/(2/3)}}}
{{{s=1.499975}}}

c) Use the same technique but with 12 terms
{{{s=(1-r^(n+1))/(1-r)}}} 
let r=1/3 
{{{s=(1-(1/3)^12)/(1-(1/3))}}}
{{{s=(1-(1/531441))/(2/3)}}}
{{{s=1.4999971}}} If it seems this sum is approaching 1.5 it is. If you go to the millionth term it will be even closer to 1.5

d)Using what we found from the two previous problems we can see that the partial sums, the pieces that make up the whole sum, are smaller than its previous term. This allows the entire series to approach 1.5