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 : Sx – 1 = S49 – Sx]
Step-by-Step Solution
Key Concept: Use the formula for the sum of the first n natural numbers, \(S_n = \frac{n(n+1)}{2}\). Express the sum of houses before \(x\) as \(S_{x-1}\) and the sum after \(x\) as \(S_{49} - S_x\). Equate the two expressions and solve the resulting quadratic equation for \(x\).
1. Write the sums using the AP‑sum formula
\[ S_{x-1}=\frac{(x-1)x}{2},\qquad S_x=\frac{x(x+1)}{2},\qquad S_{49}=\frac{49\times 50}{2}=1225. \]
2. Set the given condition
\[ S_{x-1}=S_{49}-S_x. \]
3. Substitute the expressions
\[ \frac{(x-1)x}{2}=1225-\frac{x(x+1)}{2}. \]
4. Clear the denominator (multiply by 2)
\[ x(x-1)=2450- x(x+1). \]
5. Bring all terms to one side
\[ x(x-1)+x(x+1)=2450 \]
\[ x\big[(x-1)+(x+1)\big]=2450 \]
\[ x(2x)=2450. \]
6. Simplify
\[ 2x^{2}=2450 \]
\[ x^{2}=1225. \]
7. Take the positive square root (since house numbers are positive)
\[ x=\sqrt{1225}=35. \]
8. Verification
- Sum of houses before 35: \( S_{34}=\frac{34\times35}{2}=595 \).
- Sum of houses after 35: \( S_{49}-S_{35}=1225-\frac{35\times36}{2}=1225-630=595 \).
Both sums are equal, confirming the result.
Thus, the required house number is \(x=35\).
Correct Answer: x = 35