If this is a geometric sequence, then the terms are arn for n≥0.
a=4, ar=6, ar4=20, ar5=25.
So r=6/4=3/2; r=25/20=5/4. So this is not a geometric sequence because r is not constant.
Let's assume that p(n)=a0+a1n+a2n2+a6n3. From this we can obtain no more than 4 coefficients because we are given only 4 values. We are assuming that the missing values will fit this function.
p(0)=a0=4; we can substitute for a0 for in the following equations:
p(1)=a0+a1+a2+a3=6; a1+a2+a3=2;
p(4)=a0+4a1+16a2+64a3=20, 4a1+16a2+64a3=16, a1+4a2+16a3=4;
p(5)=a0+5a1+25a2+125a3=25; 5a1+25a2+125a3=21, a1+5a2+25a3=4.2.
p(4)-p(1)=3a2+15a3=2;
p(5)-p(1)=4a2+24a3=2.2.
4(p(4)-p(1))=12a2+60a3=8;
3(p(5)-p(1))=12a2+72a3=6.6.
Therefore 12a3=-1.4, a3=-7/60.
Substituting for a3, 3a2-7/4=2, a2=5/4.
Substituting for a2 and a3:
a1+5/4-7/60=2, a1=13/15.
p(n)=4+13n/15+5n2/4-7n3/60.
p(2)=9.8, p(3)=14.7, p(6)=29.
One solution is 4,6,9.8,14.7,20,25,29. Although this is a valid solution there are other solutions, but there is insufficient data to find a different solution.