Question 1116337
ONE WAY TO START:
The recursive relation for a geometric progression,
relating term number {{{n}}} , {{{a[n]}}} to the preceding term, {{{a[n-1]}}} is
{{{a[n]=a[n-1]*r}}} , where {{{r}}} is the common ratio.
So, applying that to the fourth, fifth and sixth terms
{{{a[5]=a[4]*r=(1/2)*r}}} ,
{{{a[6]=a[5]*r=((1/2)*r)*r=(1/2)r^2}}} , and
{{{(1/2)r^2=1/8}}} --> {{{r^2=(1/8)/(1/2)}}} --> {{{r^2=1/4}}} .
Knowing that {{{r<0}}} , we conclude that {{{r=-1/2}}} ,
and can write the first 4 terms, starting from the 4th,
and dividing by {{{(-1/2)}}} (multiplying times -2)
each time to get the term before:
{{{a[4]=1/2}}} ,
{{{a[3]=(1/2)/(-1/2)=1/2)(-2)=-1}}} ,
{{{a[2]=(-1)(-2)=2}}} ,
{{{a[1]=1(-2)=-4}}} .
 
ANOTHER WAY:
The general "formula" for term number {{{n}}} , {{{a[n]}}} ,
of a GP with first term {{{a[1]}}} and common ratio {{{r}}} is
{{{a[n]=a[1]*r^(n-1)}}} .
Applying that to {{{system(a[4]=1/2,a[6]=1/8,r<0)}}} ,
{{{system(a[1]*r^3=1/2,a[4]*r^6=1/8,r<0)}}} --> {{{system(a[1]*r^3=1/2,r^2=(1/8)/(1/2)=1/4,r<0)}}} --> {{{system(a[1]*(-1/2)^3=1/2,r=-1/2)}}} --> {{{system(a[1]*(-1/8)=1/2,r=-1/2)}}} --> {{{system(a[1]=-4,r=-1/2)}}}
 
AND FROM THERE:
The "formula" for the sum of the first {{{n}}} terms, {{{S[n]}}} , of a GP is
{{{S[n]=a[1](r^n-1)/(r-1)}}}
When {{{r<1}}} ,
the term {{{r^n}}} tends to {{{0}}} as {{{n}}} tends to {{{infinity}}} ,
and the sum to infinity is
{{{S[infinity]=(-a[1])/(1-r)}}} .
In this case the sum to infinity is easy to calculate:
{{{S[infinity]}}}{{{"="}}}{{{(-(-4))/((-1/2-1))}}}{{{"="}}}{{{4/((-3/2))}}}{{{"="}}}{{{4(-2/3)}}}{{{"="}}}{{{highlight(-8/3=-2&2/3="-2.6666...")}}}
Substituting the {{{a[1]}}} and {{{r}}} values found above,
the expression for the sum of the first {{{n}}} terms becomes
{{{S[n]}}}{{{"="}}}{{{(-(-4))((-1/2)^n-1)/(-3/2)}}}{{{"="}}}{{{4(-2/3)((-1/2)^n-1)}}}{{{"="}}}{{{(8/3)(((-1)^n-2^n)/2^n)}}}{{{"="}}}{{{(2^3/3)(((-1)^n-2^n)/2^n)}}}{{{"="}}}{{{(1/2^(n-3))(((-1)^n-2^n)/3)}}}
If that is supposed to be an irreducible fraction of the form {{{K/32}}} ,
the only possibility is {{{2^(n-3)=32}}} ,
and as {{{32=2^5}}} , {{{n-3=5}}} --> {{{highlight(n=8)}}} .
 
BOTH QUESTIONS WERE ANSWERED, BUT IF YOU WANT MORE,
knowing {{{n}}} , we could try to find that denominator.
{{{K=((-1)^n-2^n)/3=((-1)^8-2^8)/3=(1-256)/2=(-255)/3=-85}}} .
So,
{{{S[infinity]}}}{{{"="}}}{{{highlight(-85/32)=(-64-21)/32=-64/32-21/32=-2-21/32=highlight(-2&21/32)}}} .