Let A_1 and A_2 be two arithmetic means and G_1,G_2 and G_3 be three geometric means of two distinct positive numbers. Then G_1^4+G_2^4+G_3^4+G_1^2…

Let the two numbers are $a,b$a,b

$\Rightarrow a,{A}_{1},{A}_{2},b$⇒a,A1​,A2​,b are in AP.

$$\Rightarrow b=a+(4-1)d$$⇒b=a+(4−1)d

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

$$\Rightarrow {A}_{1}=a+\frac{b-a}{3}=\frac{2a+b}{3}$$⇒A1​=a+3b−a​=32a+b​

$$\Rightarrow {A}_{2}=a+\frac{b-a}{3}\cdot 2=\frac{a+2b}{3}$$⇒A2​=a+3b−a​⋅2=3a+2b​

Similarly $a,{G}_{1},{G}_{2},{G}_{3},b$a,G1​,G2​,G3​,b are in GP.

$$\Rightarrow b=a{(r)}^{5-1}$$⇒b=a(r)5−1

$$\Rightarrow r={(\frac{b}{a})}^{\frac{1}{4}}$$⇒r=(ab​)41​

$$\Rightarrow {G}_{1}=a{(\frac{b}{a})}^{\frac{1}{4}}$$⇒G1​=a(ab​)41​

$$\Rightarrow {G}_{2}=a{(\frac{b}{a})}^{\frac{2}{4}}$$⇒G2​=a(ab​)42​

$$\Rightarrow {G}_{3}=a{(\frac{b}{a})}^{\frac{3}{4}}$$⇒G3​=a(ab​)43​

$$={({G}_{1})}^{4}+{({G}_{2})}^{4}+{({G}_{3})}^{4}+{({G}_{1})}^{2}\cdot {({G}_{3})}^{2}$$=(G1​)4+(G2​)4+(G3​)4+(G1​)2⋅(G3​)2

$$\Rightarrow {a}^{4}\cdot \frac{b}{a}+{a}^{4}\cdot \frac{{b}^{2}}{{a}^{2}}+{a}^{4}\cdot \frac{{b}^{3}}{{a}^{3}}+{a}^{4}\cdot \frac{{b}^{2}}{{a}^{2}}$$⇒a4⋅ab​+a4⋅a2b2​+a4⋅a3b3​+a4⋅a2b2​

$$=b{a}^{3}+{b}^{2}{a}^{2}+{b}^{3}a+{a}^{2}{b}^{2}$$=ba3+b2a2+b3a+a2b2

$$=ab({a}^{2}+{b}^{2}+2ab)=ab{(a+b)}^{2}$$=ab(a2+b2+2ab)=ab(a+b)2

$$\Rightarrow {({A}_{1}+{A}_{2})}^{2}\cdot {G}_{1}{G}_{3}={(a+b)}^{2}\cdot ab$$⇒(A1​+A2​)2⋅G1​G3​=(a+b)2⋅ab

Hence this is the correct option.