A class has 49 students who are allotted roll | vedclass

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Question

(B) Let the required roll number be $n$.According to the problem,the sum of all roll numbers smaller than $n$ is equal to the sum of all roll numbers

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(B) Let the required roll number be $n$.According to the problem,the sum of all roll numbers smaller than $n$ is equal to the sum of all roll numbers greater than $n$.This can be written as: $1 + 2 + 3 + \ldots + (n - 1) = (n + 1) + (n + 2) + \ldots + 49$.To solve this,we add the sum of numbers from $1$ to $n$ to both sides:$1 + 2 + \ldots + (n - 1) + n = (n + 1) + (n + 2) + \ldots + 49 + n$.The left side becomes the sum of the first $n$ natural numbers: $\frac{n(n + 1)}{2}$.The right side becomes the sum of the first $49$ natural numbers: $\frac{49(50)}{2}$.Thus,$2 	imes \frac{n(n + 1)}{2} = \frac{49 	imes 50}{2} + \frac{n(n + 1)}{2}$ is not the correct approach; rather,we use the property: $\sum_{i=1}^{n-1} i = \sum_{i=n+1}^{49} i$.We know $\sum_{i=1}^{n-1} i = \frac{(n-1)n}{2}$.The sum $\sum_{i=n+1}^{49} i = \sum_{i=1}^{49} i - \sum_{i=1}^{n} i = \frac{49 	imes 50}{2} - \frac{n(n+1)}{2}$.Equating the two: $\frac{n^2 - n}{2} = \frac{2450}{2} - \frac{n^2 + n}{2}$.$n^2 - n = 2450 - n^2 - n$.$2n^2 = 2450$.$n^2 = 1225$.$n = \sqrt{1225} = 35$.Therefore,the required roll number is $35$.

$A$ class has $49$ students who are allotted roll numbers from $1$ to $49$. Find a roll number $n$ such that the sum of all the roll numbers smaller than $n$ is equal to the sum of all the roll numbers greater than $n$.

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