Question

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 the value of x.
​

Answer 1

Answer:

35

Step-by-step explanation:  Using the concepts of AP:

Sum of no. of house(x - 1) preceding x is:

⇒ (x - 1)/2 [2(1) + (x - 1 - 1)1 ]

⇒ x(x - 1)/2

Remaining houses = 49 - x

Sum of no. of house(x) following x is

⇒ (49 - x)/2 [x+1 + 49]

⇒ (49 - x)(x + 50)/2

As both are equal,

x(x - 1)/2 = (49 - x)(x + 50)/2

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

x² - x = 49x + 2450 - 50x - x²

x = 35

        Method 2:

Total sum of no. of house is:

⇒ (49/2) [2(1) + (49 - 1)1 ]

⇒ (49/2) [2 + 48]    ⇒ 1225

Total sum = Sum of houses preceding x + sum of house following(including x)

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

⇒ 1225 = (x² - x + x² + x)/2

⇒ 1225 = 2x²/2

⇒ ±35 = x

As x can't be - ve, neglect x = -35.

 Required value of x is 35.

Answer 2

Given :-

The 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.

To Find :-

Value of x

Solution :-

Finding the total no. of houses

$\sf No. \; of \; house = \dfrac{49}{2} \times \bigg(2(1) + 49-1\bigg)$

$\sf No. \; of \; houses = \dfrac{49}{2}\bigg(2 + 48\bigg)$

$\sf No. \; of \; house = \dfrac{49}{2} \times 50$

$\sf No. \; of \; houses = 49 \times 25$

$\sf No. \; of \; houses = 1225$

Now,

$\sf 1225 = \dfrac{x(x-1)}{2}+ \dfrac{x}{2} \times \bigg(2+x-1\bigg)$

$\sf 1225 = \dfrac{x^2-x+x^2+x}2$

Cancelling x

$\sf 1225 = \dfrac{x^2+x^2}{2}$

$\sf 1225 \times 2 = 2x^2$

$\sf 2450= 2x^2$

$\sf \dfrac{2450}{2} = x^2$

$\sf 1225 = x^2$

$\sf 35 = x$