The correct option is B 2abc
Given that, 2a=b+c
and b,G1,G2,c are in G.P.
Therefore, G12=bG2 ....(1)
⇒G13=bG2G1
and G22=cG1 ....(2)
⇒G23=cG1G2
Multiplying (1) & (2), we get
G1G2=bc ...(3)
Therefore, G31+G32=bG2G1+cG2G1=G2G1(b+c)
⇒G31+G32=2abc ....[ From (3) ]
Ans: B