The sum of squares of three consecutive natur | vedclass

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

Question

(B) Let the three consecutive natural numbers be $(x-1)$,$x$,and $(x+1)$.According to the problem,the sum of their squares is $2030$:$(x-1)^2 + x^2 +

Vedclass · Accepted Answer

$26$

Vedclass · Answer

(B) Let the three consecutive natural numbers be $(x-1)$,$x$,and $(x+1)$.According to the problem,the sum of their squares is $2030$:$(x-1)^2 + x^2 + (x+1)^2 = 2030$Expanding the squares:$(x^2 - 2x + 1) + x^2 + (x^2 + 2x + 1) = 2030$Combining like terms:$3x^2 + 2 = 2030$Subtracting $2$ from both sides:$3x^2 = 2028$Dividing by $3$:$x^2 = 676$Taking the square root:$x = \sqrt{676} = 26$Therefore,the middle number is $26$.

The sum of squares of three consecutive natural numbers is $2030$. What is the middle number?

Similar Questions

$N$ is the largest two-digit number which,when divided by $3, 4,$ and $6,$ leaves the remainders $1, 2,$ and $4$ respectively. What is the remainder when $N$ is divided by $5$?

$(13.99)^{2} - (15.02)^{2} + (18.07)^{2} - 36.64 = ?$

$26 \times 451 - ? = 5109$

Find the square root of $(2.25)^{2} + 16 - 4(2.25) - \frac{(2.25)^{3}}{2.25 + 4} ?$

$\frac{601}{49} \times \frac{399}{81} \div \frac{29}{201} = ?$

Vedclass Test Series

Exam Paper Generator

Online Exam Module