The correct option is A c=aba+b
Let 2a=3b=6c=k
So, 2a=k
⇒k1a=2 .....(i)
Similarly k1b=3 .....(ii)
k1c=6 .....(iii)
Now, we know that 6=2×3
Putting the values of 6, 2 and 3 from (i), (ii) and (iii) we get:
k1c=k1a×k1b⇒k1c=k1a+1b⇒1c=1a+1b⇒1c=b+aab⇒c=aba+b