<s>[INST] The common difference is equal to the common ratio. Four numbers are in A.P. The first, the second and the fourth are in G.P. Find the numbers if the common difference is equal to the common ratio.
Let the terms of the A.P. be $a_1,a_1+d,a_1+2d,a_1+3d$ and the terms of the G.P. will be $a_1,a_1+d,a_1+3d.$ We know that $d=q.$ I am not sure what to do from here.
 [/INST] Let the four numbers be $a_1, a_2, a_3, a_4$. The common difference $d$ will make the terms equal to $a_1, a_1+d, a_1+2d, a_1+3d$, and the terms of the geometric sequence should be $a_1, a_1+d, a_1+3d$ as you correctly mentioned.
Since $d=r$, $a_1, a_1r, a_1r^2$ are also the terms of the geometric sequence.
So we have $a_1r=a_1+d$, $a_1r^2=a_1+3d$. Solving the first equation for $r$ gives $r=1+\frac{d}{a_1}$. Sustituting $r$ into the second equation gives $a_1(1+\frac{d}{a_1})^2=a_1+3d\Rightarrow a_1(1+\frac{2d}{a_1}+\frac{d^2}{a_1^2})=a_1+3d\Rightarrow a_1+2d+\frac{d^2}{a_1}=a_1+3d$
Simlifying gives $d^2=a_1d\Rightarrow d=0,a_1$.
A common difference of $0$ will give a common ratio of $1$ (since all $4$ terms are the same), so that's an extraneous solution.
The only other option is $d=a_1$.
In that case, substituting it back into $r=1+\frac{d}{a_1}$ gives $r=2$.
Therefore, since $d=r$, $d=r=a_1=2$, and so your only sequence is $2, 4, 6, 8$
</s>