If a, b, c are in AP, b -a, c - b and a are in GP, then a : b : c is

The correct option is A 1 : 2 : 3

a, b, c are in A.P

$\Rightarrow 2b=a+c$

Let $a=k$

$\Rightarrow c=2b-k$ and $b-a,c-b,a$ are in G.P.

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

$\Rightarrow c^2+b^2-2bc=ab-a^2$

$\Rightarrow a^2+b^2+c^2=2bc+ab$

$\Rightarrow k^2+b^2+4b^2+k^2-4bk=2b(2b-k)+bk$

$\Rightarrow 2k^2+5b^2-4bk=4b^2-bk$

$\Rightarrow 2b^2+b^2-3bk=0$

$\Rightarrow 2k^2-3bk+b^2=0$

$\Rightarrow 2k^2-2bk-bh+b^2=0$

$\Rightarrow 2k(k-b)-b(k-b)=0$

$\Rightarrow (2k-b)(k-b)=0$

$\Rightarrow b=2k$

or $b=k\times \{b\ne a\}$

$\{b\ne a$ as $b-a$ term would become $0$ & hence every term$\}$

$\Rightarrow a:b:c=k:2k:3k=1:2:3$

$\Rightarrow \left(A\right)$.