Question

The houses of a row are numbered consecutively from 1 to 50. Show that there is value of x such that the sum 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.

Read numbers carefully.

Answer 1

Let there be a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it

Let there be a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it House H

Let there be a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it House H3 ...H x−1 ,H xx1 ....H

....H 49

House number will form an A.P whose first term and the common difference is 1

House number will form an A.P whose first term and the common difference is 1Sum of n terms S=

⇒

⇒ 2

⇒ 2x−1

⇒ 2x−1

⇒ 2x−1 [x]=

⇒ 2x−1 [x]= 2

⇒ 2x−1 [x]= 249

⇒ 2x−1 [x]= 249

⇒ 2x−1 [x]= 249 [50]−

⇒ 2x−1 [x]= 249 [50]− 2

⇒ 2x−1 [x]= 249 [50]− 2x

⇒ 2x−1 [x]= 249 [50]− 2x

⇒ 2x−1 [x]= 249 [50]− 2x [x+1]

⇒ 2x−1 [x]= 249 [50]− 2x [x+1]

⇒ 2x−1 [x]= 249 [50]− 2x [x+1] [x−1+x+1]=

⇒ 2x−1 [x]= 249 [50]− 2x [x+1] [x−1+x+1]= [50]

⇒ 2x−1 [x]= 249 [50]− 2x [x+1] [x−1+x+1]= [50] [2x]=49×25

⇒ 2x−1 [x]= 249 [50]− 2x [x+1] [x−1+x+1]= [50] [2x]=49×25 =49×25

⇒ 2x−1 [x]= 249 [50]− 2x [x+1] [x−1+x+1]= [50] [2x]=49×25 =49×25⇒x=7×5=35.

Answer 2

Step-by-step explanation:

Let there be a value of x such that the sum of the number of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it

House H 1 ,H 2 ,H 3 ___.H

x−1 ,H x+1 ____H 49

House No. 1 2

3 ___x -1

x + 1.._____49

House number will form an A.P whose first term and the common difference is 1

Sum of n terms S= 2n [2a+(n−1)×d]

S x−1 =S 49 −S x⇒ 2(x−1) [2(1)+(x−1−1)1]= 249[2+48]− 2x [2(1)+(x−1)1]⇒ 2x−1 [2+(x−2)]=

249 [50]− 2x [2+x−1]⇒ 2x−1 [x]= 249

[50]− 2x [x+1]⇒ 2x [x−1+x+1]= 249 [50]

⇒ 2x [2x]=49×25

⇒x 2 =49×25

⇒x=7×5=35.

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

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