Question 969971
We need more information than that.
Is there a commonly used pattern to your sequence?
Is 3 the second term, and 17 the third term?
 
A sequence could have no rhyme or reason, such as
potato , 3 , 7, duck , 1 , chair .
However, in algebra class teachers like sequences made with some sort of pattern.
The sequences with the commonly used patterns are an arithmetic sequence or a geometric sequence.
 
If it is an arithmetic sequence, then each term is the one before plus a fixed number, {{{d}}} .
That number {{{d}}} is called the common difference, because it is the difference between any two consecutive terms.
So, term number {{{n+1}}} , {{{a[n+1]}}} ,
is related to the previous term,
term number {{{n}}}={{{a[n]}}},
by {{{a[n+1]=a[n]+d}}}<--->{{{d=a[n+1]-a[n]}}}
for all natural number (counting number) values of {{{n}}} .
If 3 and 17 are the 2nd and 3rd terms respectively,
then the common difference between consecutive terms is
{{{d-17-3=14}}} .
Each term is {{{14}}} more than the one before.
{{{a[2]=3}}} , {{{a[3]=17}}} , {{{d=14}}} , and {{{a[1]}}} is the first term that we want to find.
From {{{a[n+1]=a[n]+d}}} , for {{{n=1}}} , we get
{{{a[1+1]=a[1]+d}}}--->{{{a[2]=a[1]+14}}}--->{{{3=a[1]+14}}}
so {{{a[1]+14=3}}}--->{{{a[1]=3-14}}}--->{{{a[1]=-11}}}
 
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 the 2nd and 3rd terms respectively,
then the common ratio between consecutive terms is
{{{r=17/3}}} .
Each term is {{{17/3}}} times the one before.
{{{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}}}