Question 52

The sum of the square of 3 consecutive positive numbers is 365. The sum of the numbers is

Solution

Let the 3 consecutive positive numbers = $$(x-1),(x),(x+1)$$

According to ques, => $$(x-1)^2+(x)^2(x+1)^2=365$$

=> $$(x^2-2x+1)+(x^2)+(x^2+2x+1)=365$$

=> $$3x^2+2=365$$

=> $$3x^2=365-2=363$$

=> $$x^2=\frac{363}{3}=121$$

=> $$x=\sqrt{121}=11$$

$$\therefore$$ Sum of numbers = $$(x-1)+(x)+(x+1)=3x$$

= $$3\times11=33$$

=> Ans - (B)

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