From natural numbers 1 to 288, find the numbe | vedclass

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

Question

(C) Let the required number be $x.$ The sum of natural numbers smaller than $x$ is given by the sum of integers from $1$ to $x-1.$This sum is $S_1 = \

Vedclass · Accepted Answer

$204$

Vedclass · Answer

(C) Let the required number be $x.$ The sum of natural numbers smaller than $x$ is given by the sum of integers from $1$ to $x-1.$This sum is $S_1 = \frac{(x-1)x}{2}.$The sum of natural numbers greater than $x$ up to $288$ is the sum of integers from $x+1$ to $288.$This sum is $S_2 = \sum_{i=1}^{288} i - \sum_{i=1}^{x} i = \frac{288 	imes 289}{2} - \frac{x(x+1)}{2}.$According to the problem,$S_1 = S_2.$$\frac{x^2 - x}{2} = \frac{288 	imes 289}{2} - \frac{x^2 + x}{2}.$Multiplying by $2,$ we get $x^2 - x = 288 	imes 289 - x^2 - x.$$2x^2 = 288 	imes 289.$$x^2 = 144 	imes 289.$$x = \sqrt{144 	imes 289} = 12 	imes 17 = 204.$Thus,the required number is $204.$

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 $288.$

Similar Questions

For a given $A.P.$,$a=2$ and $d=3$. Then,$S_{30} = \dots$

The sum of the first $20$ terms of the $A.P.$ $5, 11, 17, 23, \ldots$ is:

The $n^{th}$ term of an $A.P.$ is given by $T_{n} = 4n + 7$. Find the sum of $n$ terms of this $A.P.$

Show that $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ form an $AP$ where $a_{n}$ is defined as $a_{n}=3+4 n$. Also,find the sum of the first $15$ terms.

Find the sum of first $n$ terms of the $A.P.$ $6, 2, -2, \ldots$

Vedclass Test Series

Exam Paper Generator

Online Exam Module