A class has 49 students who are allotted roll | vedclass

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

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 $n$.

Similar Questions

For an $A.P.$,$a=11$ and $d=7$. Find the sum of the first $40$ terms of the $A.P.$

From natural numbers $1$ to $288,$ find the number $x$ such that the sum of all natural numbers smaller than $x$ is equal to the sum of all natural numbers greater than $x$ but less than or equal to $288.$

With respect to the usual notations of an $A.P.$,if $T_{n} = 28$,$S_{n} = 144$,and $n = 9$,find the first term $a$.

For an $A.P.$,the $12^{th}$ term is $-13$ and the sum of its first $4$ terms is $24$. Find the sum of the first $10$ terms of the $A.P.$

For a given $A.P.$,$S_{10} = 50$ and $a = 0.5$. Then,$d = \ldots \ldots \ldots . .$

Vedclass Test Series

Exam Paper Generator

Online Exam Module