```Question 53621
Geometric Sequence: {1,1/2,1/4,1/8...}
a)r=a(sub n+1)/a(sub n)  in some books r=a(sub n)/a(sub n-1).  It's the same thing, divide the 2nd term by the 1st, the 3rd by the 2nd, ect.
r=(1/2)/(1)=1/2
---
r=(1/4)/(1/2)
r=(1/4)*(2/1)
r=2/4=1/2
---
r=(1/8)/(1/4)
r=(1/8)*(4/1)
r=4/8=1/2
---
Notice that in geometric sequences, you will get the same ratio with any two consecutive terms, in this case:r=1/2.
-------------------------------------------------------------------------------
b)The sum of the first n terms is:
{{{S(sub n)=a(sub 1)*(1-r^n)/(1-r)}}}
In this case you want the sum of the first 10 terms.
n=10
r=1/2
The first term, a(sub 1)=1
{{{S(sub 10)=1*(1-(1/2)^10)/(1-1/2)}}}
{{{S(sub 10)=(1-1/1024)/(1-1/2)}}}
{{{S(sub 10)=(1024/1024-1/1024)/(2/2-1/2)}}}
{{{S(sub 10)=(1023/1024)/(1/2)}}}
{{{S(sub 10)=(1023/1024)*(2/1)}}}
{{{S(sub 10)=1023/512}}}
S(sub 10)=1.9980
-------------------------------------------------------------------------------
c)  This time
n=12
r=1/2
a(sub 1)=1
{{{S(sub n)=a(sub 1)*(1-r^n)/(1-r)}}}
{{{S(sub 12)=1*(1-(1/2)^12)/(1-1/2)}}}
{{{S(sub 12)=(1-1/4096)/(1-1/2)}}}
{{{S(sub 12)=(4096/4096-1/4096)/(2/2-1/2)}}}
{{{S(sub 12)=(4095/4096)/(1/2)}}}
{{{S(sub 12)=(4095/4096)*(2/1)}}}
{{{S(sub 12)=4095/2048}}}
S(sub 12)=1.9995
Happy Calculating!```