The correct option is C aba+b
Let 2a=3b=6c=k
So, 2a=k
⇒k1a=2 ... (i)
Similarly, k1b=3 ... (ii)
and 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