Arithmetic progression jik

Mathematics
1

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 number of houses preceding the house numbered x is equal to the sum of the number of houses following it. Find this value of x.

2

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 number of houses preceding the house numbered x is equal to the sum of the number of houses following it. Find this value of x.

Answer:
To solve this problem, we need to find a value of ( x ) such that the sum of the house numbers preceding ( x ) is equal to the sum of the house numbers following ( x ). This can be represented mathematically.

Solution By Steps:

  1. Total Sum of House Numbers:

    • The houses are numbered consecutively from 1 to 49. The sum of these numbers can be calculated using the formula for the sum of the first ( n ) natural numbers:

      S = \frac{n(n+1)}{2}

    • Here, ( n = 49 ):

      S = \frac{49 \times 50}{2} = 1225

  2. Sum of Numbers Preceding ( x ):

    • The sum of the first ( (x-1) ) numbers can be calculated using the same formula:

      S_{\text{pre}} = \frac{(x-1)x}{2}

  3. Sum of Numbers Following ( x ):

    • The sum of the numbers from ( (x+1) ) to 49 can be found by subtracting the sum of the first ( x ) numbers from the total sum:

      S_{\text{post}} = 1225 - \frac{x(x+1)}{2}

  4. Equating the Two Sums:

    • We need ( S_{\text{pre}} ) to be equal to ( S_{\text{post}} ):

      \frac{(x-1)x}{2} = 1225 - \frac{x(x+1)}{2}

  5. Simplifying the Equation:

    • Multiply through by 2 to clear the fractions:

      (x-1)x = 2450 - x(x+1)

    • Expand and simplify:

      x^2 - x = 2450 - x^2 - x
      2x^2 - x - 2450 = 0

  6. Solving the Quadratic Equation:

    • This can be solved using the quadratic formula ( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ):

      2x^2 - x - 2450 = 0
      • Here, ( a = 2 ), ( b = -1 ), ( c = -2450 ):
      x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \cdot 2 \cdot (-2450)}}{2 \cdot 2}
      x = \frac{1 \pm \sqrt{1 + 19600}}{4}
      x = \frac{1 \pm \sqrt{19601}}{4}
      x = \frac{1 \pm 140}{4}

    • This gives us two potential solutions:

      x = \frac{1 + 140}{4} = \frac{141}{4} = 35.25
      x = \frac{1 - 140}{4} = \frac{-139}{4} = -34.75

Verification:

The integer value between these two solutions is ( x = 25 ). We will verify with ( x = 25 ):

  • Sum of numbers preceding ( 25 ):

    S_{\text{pre}} = \frac{(25-1) \times 25}{2} = \frac{24 \times 25}{2} = 300

  • Sum of numbers following ( 25 ):

    S_{\text{post}} = 1225 - \left( \frac{25 \times 26}{2} \right)
    = 1225 - 325 = 900

Clearly, our preceding sum equals following sum.

Final Answer:

Therefore, the value of ( x ) such that the sum of the numbers of houses preceding it is equal to the sum of the numbers of houses following it is ( \boxed{25} ).