If the square of the sum of three positive co | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Let the three consecutive natural numbers be $(x-1)$,$x$,and $(x+1)$.According to the problem,the square of their sum exceeds the sum of their squ

Vedclass · Accepted Answer

$8$

Vedclass · Answer

(D) Let the three consecutive natural numbers be $(x-1)$,$x$,and $(x+1)$.According to the problem,the square of their sum exceeds the sum of their squares by $292$.The sum of the numbers is $(x-1) + x + (x+1) = 3x$.The square of the sum is $(3x)^2 = 9x^2$.The sum of their squares is $(x-1)^2 + x^2 + (x+1)^2 = (x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) = 3x^2 + 2$.Given that $9x^2 - (3x^2 + 2) = 292$.$6x^2 - 2 = 292$.$6x^2 = 294$.$x^2 = 49$.$x = 7$.The three numbers are $(7-1)$,$7$,and $(7+1)$,which are $6$,$7$,and $8$.The largest of the three numbers is $8$.

If the square of the sum of three positive consecutive natural numbers exceeds the sum of their squares by $292$,then what is the largest of the three numbers?

Similar Questions

What will come in the place of the question mark $(?)$ in the following expression?
$\sqrt{15^{2} \times 12 \div 9 - 125 + 21} = ?$

Which of the following statement$(s)$ is/are $TRUE$?
$I.$ $\sqrt{11}+\sqrt{7} < \sqrt{10}+\sqrt{8}$
$II.$ $\sqrt{17}+\sqrt{11} > \sqrt{15}+\sqrt{13}$

What will come in the place of the question mark $(?)$ in the following equation?
$(0.64)^{4} \div (0.512)^{3} \times (0.8)^{4} = (0.8)^{?+3}$

$47 \frac{1}{17} \div 1 \frac{49}{51} + 23 \frac{5}{7} + ? = 67 \frac{4}{9}$

The simplification of $1+\frac{1}{2+\frac{1}{1-\frac{1}{3}}}$ yields the result

Vedclass Test Series

Exam Paper Generator

Online Exam Module