Question 1040859
In geometric progression, each term is equal to the one before times a certain number called the common ratio, {{{r}}} .
If we call the first term {{{b[1]}}},
everything else can be calculated from {{{b[1]}}} and {{{r}}} .
The {{{n^th}}} term is {{{b[n]=b[1]*r^(n-1)}}} ,
and the sum of the first {{{n}}} terms is
{{{S[n]=b[1]*((r^n-1)/(r-1))}}} .
If we knew {{{b[1]}}} and {{{r}}} ,
we could calculate
the sum of the first {{{4}}} terms as
{{{S[4]=b[1]*((r^4-1)/(r-1))}}} ;
the sum of the first {{{8}}} terms as
{{{S[8]=b[1]*((r^8-1)/(r-1))}}} ;
the {{{9^th}}} term as
{{{b[7]=b[1]*r^(7-1)=b[1]*r^6}}} , and
the {{{9^th}}} term as
{{{b[9]=b[1]*r^(9-1)=b[1]*r^8}}} .
The problem tells us that
"the sum of the first eight terms is seventeen times the sum of the first four terms" ,
so {{{S[8]=17*S[4]}}} or {{{b[1]*((r^8-1)/(r-1))=17*b[1]*((r^4-1)/(r-1))}}} , and
"the seventh term is {{{960}}} " ,
so {{{b[7]=960}}} or {{{b[1]*r^6=960}}} .
From the two equations above, we can find {{{b[1]}}} and {{{r}}} ,
and from {{{b[1]}}} and {{{r}}} ,
we can find {{{b[9]=b[1]*r^8}}} .
Alternately, {{{b[9]=b[8]*r=b[7]*r^2}}} .
 
{{{b[1]*((r^8-1)/(r-1))=17*b[1]*((r^4-1)/(r-1))}}}
{{{(r^8-1)=17*(r^4-1)}}}
{{{(r^8-1)/(r^4-1)=17}}}
{{{r^4+1=17}}}
{{{r^4=17-1}}}
{{{r^4=16}}}
{{{r=root(4,16)}}}
{{{highlight(r=2)}}}
Then {{{b[9]=b[7]*r^2=960*2^2=960*4=3840}}}
So, {{{b[9]=highlight(3840)}}} .