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 \(x\).
Hint: \(S_{x-1} = S_{49} - S_x\).
35
Step 1: Use the sum formula for natural numbers.
The house numbers form an AP: \(1, 2, 3, \ldots, 49\).
Sum of first \(n\) natural numbers is:
\[S_n = \frac{n(n + 1)}{2}\]
Step 2: Translate the condition into an equation.
Let the required house number be \(x\).
• Sum of the house numbers before house \(x\):
\[S_{x-1} = 1 + 2 + \cdots + (x - 1) = \frac{(x - 1)x}{2}\]
• Sum of the house numbers after house \(x\):
\[S_{49} - S_x\]
Given condition: these two sums are equal:
\[S_{x-1} = S_{49} - S_x\]
Step 3: Substitute the formulas.
First, find \(S_{49}\):
\[S_{49} = \frac{49 \cdot 50}{2} = 1225\]
Also,
\[S_x = \frac{x(x + 1)}{2}\]
So the condition becomes:
\[\frac{(x - 1)x}{2} = 1225 - \frac{x(x + 1)}{2}\]
Step 4: Clear the denominator and simplify.
Multiply both sides by 2:
\[x(x - 1) = 2450 - x(x + 1)\]
Expand both sides:
Left: \(x^2 - x\)
Right: \(2450 - (x^2 + x) = 2450 - x^2 - x\)
So the equation is:
\[x^2 - x = 2450 - x^2 - x\]
Bring all terms to one side:
\[x^2 - x + x^2 + x - 2450 = 0\]
\[2x^2 - 2450 = 0\]
Step 5: Solve for \(x\).
\[2x^2 = 2450 \Rightarrow x^2 = 1225\]
\[x = \pm 35\]
But \(x\) is a house number from 1 to 49, so \(x = 35\).
Step 6: Quick check.
Sum of houses before 35:
\[S_{34} = \frac{34 \cdot 35}{2} = 595\]
Sum of houses after 35:
\[S_{49} - S_{35} = 1225 - \frac{35 \cdot 36}{2} = 1225 - 630 = 595\]
Both sums are equal.
Conclusion: The required house number is \(x = 35\).