The answer depends on what you know about the sequence [tex]a_n[/tex]...
If [tex]a_n[/tex] is geometric, then
[tex]a_n=ra_{n-1}[/tex]
for some fixed number r. Using this recursive rule, we get
[tex]a_{n-1}=ra_{n-2}\implies a_n=r^2a_{n-2}[/tex]
[tex]a_{n-2}=ra_{n-3}\implies a_n=r^3a_{n-3}[/tex]
and so on, down to the 1st term in the sequence:
[tex]a_n=r^{n-1}a_1[/tex]
This means
[tex]a_3=81=r^2a_1[/tex]
[tex]a_6=648=r^5a_1[/tex]
so that
[tex]\dfrac{648}{81}=\dfrac{r^5a_1}{r^2a_1}\implies 8=r^3\implies r=2[/tex]
Now solve for [tex]a_1[/tex]:
[tex]81=2^2a_1\implies a_1=\boxed{\dfrac{81}4}[/tex]
If instead [tex]a_n[/tex] is arithmetic, then
[tex]a_n=a_{n-1}+d[/tex]
for some fixed number d. Similarly, we can get the n-th number in the sequence in terms of the 1st one:
[tex]a_{n-1}=a_{n-2}+d\implies a_n=a_{n-2}+2d[/tex]
[tex]a_{n-2}=a_{n-3}+d\implies a_n=a_{n-3}+3d[/tex]
and so on, down to
[tex]a_n=a_1+(n-1)d[/tex]
Then
[tex]a_3=81=a_1+2d[/tex]
[tex]a_6=648=a_1+5d[/tex]
Solve for d :
[tex](a_1+5d)-(a_1+2d)=648-81\implies 3d=567\implies d=189[/tex]
Now solve for [tex]a_1[/tex]:
[tex]81=a_1+2\cdot189\implies a_1=\boxed{-297}[/tex]
