Odd integers between 2 and 22Â = 3, 5, 7,......, 21
This is an A.P. with first term = $$a=3$$ and $$d=2$$
Let number of terms = $$n$$
=> Last term = $$a+(n-1)d$$
=> $$3+(n-1)2=21$$
=> $$(n-1)2=21-3=18$$
=> $$(n-1)=\frac{18}{2}=9$$
=> $$n=9+1=10$$
Sum of series = $$\frac{n}{2}(a+l)$$
=Â $$\frac{10}{2}(3+21)$$
= $$5\times24=120$$
$$\therefore$$ Average = $$\frac{120}{10}=12$$
=> Ans - (D)
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