The houses in a row numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.

Given: The houses in a row numbered consecutively from 1 to 49

To do: To show that there exists a value of X such that sum of numbers of houses preceding the house numbered X is equal to sum of the numbers of houses following X.

Solution:

Let there be a value of $x$ such that the sum of the numbers of the houses preceding the house numbered $x$ is equal to the sum of the numbers of the houses following it.

That is, $1\ +\ 2\ +\ 3\ +\ .\ .\ .\ .\ +\ ( x\ –\ 1)=( x+1)+( x+2)+\ .\ .\ .\ .\ .\ + 49$

$\therefore 1+2+3+\ .\ .\ .\ .\ +( x–1)=[1+2+\ ....+x+( x–1)+....+49] \ –\ ( 1+2+3+.\ .\ .\ .\ +x$

As known sum of the n terms of the A.P. $S_{n} =\frac{n}{2}( a+l)$   [a is the first term and l is the last term of the given A.P.]

$\frac{x-1}{2}( 1+x-1) =\frac{49}{2}( 1+49) -\frac{x}{2}( a+l)$

$\therefore  x( x\ –\ 1) \ =\ 49\ \times \ 50\ –\ x( 1\ +\ x)$

$\therefore  x( x\ –\ 1) \ +\ x( 1\ +\ x) \ =\ 49\ \times \ 50$

$\therefore  x( x-1+1+x) =49\times 50$

$2x^{2} = 49\times50$

$\therefore x^{2} =49\times 25$

$\Rightarrow x=\sqrt{49\times 25}$

$\therefore x=7\times 5=35$

Since $x$ is not a fraction, The value of $x$ satisfying the given condition exists and is equal to 35.

 

Updated on: 10-Oct-2022

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