Question 76

Find the average of first 20 multiples of 9.

Solution

First 20 multiples of 9 = $$9,18,27,......,171,180$$

=> Sum = $$9(1+2+3+.....+19+20)$$

= $$9\times\frac{20(20+1)}{2}$$ $$[\because$$ Sum of first $$n$$ integers $$=\frac{n(n+1)}{2}]$$

= $$9\times10\times21=1890$$

=> Average = $$\frac{1890}{20}=94.5$$

=> Ans - (D)

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