The sum of the square of 3 consecutive positive numbers is 365. The sum of the numbers is
Let the 3 consecutive positive numbers = $$(x-1),(x),(x+1)$$
According to ques, =>Â $$(x-1)^2+(x)^2(x+1)^2=365$$
=> $$(x^2-2x+1)+(x^2)+(x^2+2x+1)=365$$
=> $$3x^2+2=365$$
=> $$3x^2=365-2=363$$
=> $$x^2=\frac{363}{3}=121$$
=> $$x=\sqrt{121}=11$$
$$\therefore$$ Sum of numbers =Â $$(x-1)+(x)+(x+1)=3x$$
= $$3\times11=33$$
=> Ans - (B)
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