Multiples of 3 are : 15,18,21,.....,102,105
The above series is an A.P. with first term =Â $$a=15$$ and common difference = $$d=3$$ and last term = $$l=105$$
Let number of terms be $$n$$
=> Number of terms = $$l=a+(n-1)d$$
=> $$15+(n-1)3=105$$
=> $$(n-1)3=105-15=90$$
=> $$n-1=\frac{90}{3}=30$$
=> $$n=31$$
=> Ans - (B)
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