The common difference is equal to the common ratio.
Matthew Harrington 
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.
$\endgroup$4 Answers
$\begingroup$Write $\dfrac {a_1+d}{a_1} = \dfrac{a_1+3d}{a_1+d} = d$.
Now we also have $\dfrac {d}{a_1} = \dfrac {2d}{2a_1} =\dfrac{2d}{a_1+d} = d-1$.
This gives $a_1 = d$, and from $\dfrac {a_1+d}{a_1}=d$ we have $a_1=d=2$.
$\endgroup$ 4 $\begingroup$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 $2, 4, 6, 8$
$\endgroup$ 4 $\begingroup$Let the four numbers in A.P. be $a-3d,a-d,a+d,a+3d$. Common difference is $2d$. Again 1st,2nd,4th are in G.P. with $2d$ as common ratio i.e. ${a-d\over a-3d}={a+3d\over a-d}=2d$.
Simplifying $(a-d)^2=(a-3d)(a+3d)$ we get $a=5d$. Putting this into ${a-d\over a-3d}=2d$ we get $d=1\implies a=5$.
Hence the four numbers are $2,4,6$ and $8$.
$\endgroup$ 0 $\begingroup$Since the common difference $ \ d \ $ in the arithmetic progression is equal to the common ratio $ \ r \ $ in the geometric progression, the geometric progression can be written as$$ a_1 \ \ \ , \ \ \ a_2 \ \ = \ \ a_1 + d \ \ = \ \ a_1·d \ \ \ , \ \ \ a_4 \ \ = \ \ a_2 \ + \ 2·d \ \ = \ \ a_1·d \ + \ 2·d \ \ = \ \ a_1·d^2 \ \ . $$From this last equation, we have $ \ a_1·d^2 - (a_1 + 2)·d \ = \ 0 \ \Rightarrow \ d·( \ a_1·d - [a_1 + 2] \ ) \ = \ 0 \ \ . $ Since having $ \ d = 0 \ $ would also make $ \ r = 0 \ \ , $ it must be that $ \ a_1·d \ = \ a_1 + 2 \ \ . $
That means the arithmetic sequence is$$ a_1 \ \ \ , \ \ \ a_2 \ = \ a_1 + 2 \ \ \ , \ \ \ a_3 \ = \ a_1 + 2 + d \ \ \ , \ \ \ a_4 \ = \ a_1 + 2 + 2·d \ \ , $$which makes it evident that $ \ d = 2 \ $ and the sequence is $ \ a_1 \ , \ a_1 + 2 \ , \ a_1 + 4 \ , \ a_1 + 6 \ \ . $ But the geometric sequence then requires that $ \ a_1 + 2 \ = \ 2·a_1 \ $ and $ \ a_1 + 6 \ = \ 4·a_1 \ \ , $ either equation of which can be solved to find that $ \ a_1 \ = \ 2 \ \ . $ The progression is thus $ \ 2 \ , \ 4 \ , \ 6 \ , \ 8 \ \ . $
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