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The houses of a row are numbered consecutively from 1 to 49. Show that there is 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. Find this value of $x$.
Given:
The houses of a row are numbered consecutively from 1 to 49.
To do:
We have to show that there is 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 and find this value of $x$.
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.
This implies,
$1\ +\ 2\ +\ 3\ +\ .\ .\ .\ .\ +\ ( x\ –\ 1)=( x+1)+( x+2)+\ .\ .\ .\ .\ .\ + 49$
$\therefore 1+2+3+\ .\ .\ .\ .\ +( x–1)=[1+2+\ ....+x+( x–1)+....+49] \ –\ ( 1+2+3+.\ .\ .\ .\ +x$
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.
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