Relate separated arithmetic terms to a geometric progression

Write the AP as $T_n=a+(n-1)d$, $d\ne0$, and the GP as $u_n=br^{n-1}$. Then $a+d=b$, $a+4d=br$, $a+9d=br^2$. Thus $(a+9d)(a+d)=(a+4d)^2$, so $d(2a-7d)=0$. Rejecting $d=0$ gives $a=7d/2$ and $r=(a+4d)/(a+d)=5/3$.

$b(r-1)=3d$ and $br(r-1)=5d$. Since $d\ne0$, the first difference is nonzero, so division is valid and $r=5/3$.

Use $d=(br-b)/3$ in $br^2-br=5d$. With $b\ne0$, $3r^2-8r+5=(3r-5)(r-1)=0$. The root $r=1$ would give $3d=0$, so reject it and retain $r=5/3$.

From $a=7$ and $b=a+d=9$, $d=2$. The difference is $D_n=9(5/3)^{n-1}-2n-5$. It increases for $n\ge1$, since $D_{n+1}-D_n=6(5/3)^{n-1}-2>0$.

Thus the least n with $D_n\ge1000$ is $11$.