Question

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

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`]

Answer 1

Let there be 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.

That is, 1 + 2 + 3 ... + (x - 1) = (x + 1) + (x + 2) + ... + 49

∴ 1 + 2 + 3 + ... + (x - 1)

= [1 + 2 + ... + x + (x + 1) + ... + 49] - (1 + 2 + 3 + ... + x)

∴ `x - 1/2 [1 + x - 1] = 49/2 [1 + 49] - x/2 [1 + x]`

∴ x(x - 1) = 49 × 50 - x(1 + x)

∴ x(x - 1) + x(1 + x) = 49 × 50

∴ x² - x + x + x²

= 49 × 50

∴ x² = 49 × 25

∴ x = 7 × 5

∴ x = 35

Since x is not a fraction, the value of x satisfying the given condition exists and is equal to 35.

Answer 2

Let there be 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.

a = 1 and d = 2 - 1 = 1 and n = 49

∴ `S_(x-1)` = S₄₉ - S_x

⇒ `(x - 1)/2[2a + (x - 2)d]`

= `49/2[2a + (49 - 1)d] - x/2[2a + (x - 1) xx d]`

⇒ `(x - 1)/2[2 xx 1 + (x - 2) xx 1]`

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

⇒ (x - 1)(x)

= `49(50) - x(x + 1)`

⇒ x² - x

= 2450 - x² - x

⇒ 2x² = 2450

⇒ x² = `2450/2`

⇒ x² = 1225

⇒ x = ±`sqrt1225`

⇒ x = ± 35

Since x is not a fraction, the value of x satisfying the given condition exists and is equal to 35.