If a, b, c are in A.P. and a, b, d are in G.P.,

$a, b$ and $c$ are in A.P.

$\therefore 2 b=a+c \quad \ldots \ldots .(\mathrm{i})$

Also, $a, b$ and $d$ are in G.P.

$\therefore b^{2}=a d \quad \ldots \ldots$ (ii)

Now, $(a-b)^{2}$

$=a^{2}-2 a b+b^{2}$

$=a^{2}-a(a+c)+a d$       [ Using (1) and (2) ]

$=a d-a c$

$=a(d-c)$

$\Rightarrow(a-b)^{2}=a(d-c)$

Therefore, $a,(a-b)$ and $(d-c)$ are in G.P.