Let the two numbers are a,b
⇒a,A1,A2,b are in AP.
⇒b=a+(4−1)d
⇒d=3b−a
⇒A1=a+3b−a=32a+b
⇒A2=a+3b−a⋅2=3a+2b
Similarly a,G1,G2,G3,b are in GP.
⇒b=a(r)5−1
⇒r=(ab)41
⇒G1=a(ab)41
⇒G2=a(ab)42
⇒G3=a(ab)43
=(G1)4+(G2)4+(G3)4+(G1)2⋅(G3)2
⇒a4⋅ab+a4⋅a2b2+a4⋅a3b3+a4⋅a2b2
=ba3+b2a2+b3a+a2b2
=ab(a2+b2+2ab)=ab(a+b)2
⇒(A1+A2)2⋅G1G3=(a+b)2⋅ab
Hence this is the correct option.