Question 969778
We need more information than that.
Is 3 the second term, and 17 the third term?
 
If it is a geometric sequence, then each term is the one before times a fixed number, {{{r}}} .
That number {{{r}}} is called the common ratio, because it is the ratio between any two consecutive terms.
So, term number {{{n+1}}} , {{{b[n+1]}}} ,
is related to the previous term,
term number {{{n}}}={{{b[n]}}},
by {{{b[n+1]=b[n]*r}}}<--->{{{r=b[n+1]/b[n]}}}
for all natural number (counting number) values of {{{n}}} .

IF 3 and 17 are consecutive terms,
then the common ratio between consecutive terms is
{{{r=17/3}}} .
Each term is {{{17/3}}} times the one before.
IF 3 and 17 are the 2nd and 3rd terms respectively,
then {{{b[2]=3}}} , {{{b[3]=17}}} , {{{r=17/3}}} , and {{{b[1]}}} is the first term that we want to find.
From {{{b[n+1]=b[n]*r}}} , for {{{n=1}}} , we get
{{{b[1+1]=b[1]*r}}}--->{{{b[2]=b[1](17/3)}}}--->{{{3=b[1](17/3)}}}
so --->{{{b[1](17/3)=3}}}--->{{{b[1]=3/((17/3))}}}--->{{{b[1]=3(3/17)}}}--->{{{b[1]=9/17}}}