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(A) The house numbers are $1, 2, 3, \ldots, 49$. This forms an Arithmetic Progression $(A.P.)$ with first term $a = 1$ and common difference $d = 1$.L

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(A) The house numbers are $1, 2, 3, \ldots, 49$. This forms an Arithmetic Progression $(A.P.)$ with first term $a = 1$ and common difference $d = 1$.Let the house number be $x$. The sum of the house numbers preceding $x$ is the sum of the first $(x-1)$ terms: $S_{x-1} = \frac{(x-1)}{2}[2a + (x-1-1)d] = \frac{x-1}{2}[2(1) + (x-2)(1)] = \frac{x(x-1)}{2}$.The sum of the house numbers following $x$ is the total sum of $49$ houses minus the sum of the first $x$ houses: $S_{49} - S_x = \frac{49}{2}[2(1) + (49-1)(1)] - \frac{x}{2}[2(1) + (x-1)(1)] = \frac{49 	imes 50}{2} - \frac{x(x+1)}{2} = 1225 - \frac{x(x+1)}{2}$.Equating the two sums: $\frac{x(x-1)}{2} = 1225 - \frac{x(x+1)}{2}$.Multiplying by $2$: $x^2 - x = 2450 - (x^2 + x)$.$x^2 - x = 2450 - x^2 - x$.$2x^2 = 2450$.$x^2 = 1225$.$x = \sqrt{1225} = 35$ (since $x$ must be positive).Thus,the value of $x$ is $35$.

The houses of a row are numbered consecutively from $1$ to $49$. Show that there is 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. Find this value of $x$.

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