Question

The houses in a row are numbered consecutively from 1 to 49. Show that there exists a value of x such that the sum of numbers of houses preceding the house numbered x is equal to sum of the number of houses following x.

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Answer 1

Number of the houses as x =35

Explanation:

The number of houses is 1, 2, 3, ..., 49

By observation, the numbers of houses are in an A.P.

Hence

First-term, a = 1

Common difference, d = 1

Let us assume that the number of xth house can be expressed as below:

We know that sum of n terms in an A.P. is given by the formula Sₙ = n/2 [2a + (n - 1) d]

Sum of number of houses preceding xth house = Sₓ ₋ ₁

Sₓ ₋ ₁ = (x - 1) / 2 [2a + ((x - 1) - 1)d]

= (x - 1) / 2 [2 × 1 + ( x - 2) × 1]

= (x - 1) / 2 [2 + x - 2]

= [x (x - 1)] / 2 ---------- (1)

By the given information we know that, sum of number of houses following xth house = S₄₉ - Sₓ

S₄₉ - Sₓ = 49 / 2 [2 × 1 + (49 - 1) × 1] - x / 2 [2 × 1 + (x - 1) × 1]

= 49 / 2 [2 + 48] - x / 2 (2 + x - 1)

= (49 / 2) × 50 - (x / 2) (x + 1)

= 1225 - [x (x + 1)] / 2 ---------- (2)

It is given that these sums are equal, that is equation (1) = equation(2)

x (x - 1) / 2 = 1225 - x (x + 1) / 2

x² / 2 - x / 2 = 1225 - x² / 2 - x / 2

On solving further we get,

x² = 1225

x = ± 35

As the number of houses cannot be a negative number, we consider the number of houses as x = 35

Answer 2

35 is the correct answer

Explanation:

Given that the houses in a row are numbered consecutively from 1 to 49. We need to show that there exists a value of x such that the sum of numbers of houses preceding the house numbered x is equal to sum of the number of houses following x. Making it more simple for you; sum of (1, 2, 3, 4.... upto x - 1) is equal to sum of (x + 1), (x + 2), (x + 3).... upto 49.

→ Sum of (x - 1) terms = Sum of (49 - x) terms

"x" is the number of houses required. The sum of houses are in AP where the value of a is 1 and d is also 1; where 'a' is first term and 'd' is common difference. We know that sum of nth term of an AP is given by: T(n) = a + (n - 1)d and upto nth term by S(n) = n/2 (2a + (n - 1)d). And also by T(n) = S(n) - S(n - 1).

As per given problem:

→ (x - 1)/2 (Sum of first and last term) = (49 - x)/2 (Sum of first and last term)

→ (x - 1)(Sum of first and last term) = (49 - x) (Sum of first and last term)

Where first term and last term in LHS is 1 and (x - 1) & in RHS is (x + 1) and 49. Substitute them.

→ (x - 1) (1 + (x - 1)) = (49 - x) ((x + 1) + 49)

→ (x - 1) (1 + x - 1) = (49 - x) (x + 1 + 49)

→ (x - 1) (x) = (49 - x) (x + 50)

→ x² - x = 49x + 2450 - x² - 50x

On solving we get,

→ 2x² - 2450 = 0

→ 2x² = 2450

→ x² = 1225

→ x = ± 35

The number of houses is a positive integer. So negative value neglected. We left with x = 35. Therefore, the value of x is 35.