Let the two numbers are $a, b$

$\Rightarrow a,A_1,A_2,b$ are in AP.

$\Rightarrow b=a+\left(4-1\right)d$

$\Rightarrow d=\frac{b-a}{3}$

$\Rightarrow A_1=a+\frac{b-a}{3}=\frac{2a+b}{3}$

$\Rightarrow A_2=a+\frac{b-a}{3}·2=\frac{a+2b}{3}$

Similarly $a,G_1,G_2,G_3,b$ are in GP.

$\Rightarrow b=a{\left(r\right)}^{5-1}$

$\Rightarrow r={\left(\frac{b}{a}\right)}^{\frac{1}{4}}$

$\Rightarrow G_1=a{\left(\frac{b}{a}\right)}^{\frac{1}{4}}$

$\Rightarrow G_2=a{\left(\frac{b}{a}\right)}^{\frac{2}{4}}$

$\Rightarrow G_3=a{\left(\frac{b}{a}\right)}^{\frac{3}{4}}$

$={\left(G_1\right)}^4+{\left(G_2\right)}^4+{\left(G_3\right)}^4+{\left(G_1\right)}^2·{\left(G_3\right)}^2$

$\Rightarrow a^4·\frac{b}{a}+a^4·\frac{b^2}{a^2}+a^4·\frac{b^3}{a^3}+a^4·\frac{b^2}{a^2}$

$=b{a}^3+b^2{a}^2+b^{3}a+a^2{b}^2$

$=ab\left(a^2+b^2+2ab\right)=ab{\left(a+b\right)}^2$

$\Rightarrow {\left(A_1+A_2\right)}^2·G_1{G}_3={\left(a+b\right)}^2·ab$

Hence this is the correct option.