Question

The houses of a row in a colony 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

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=

2

n

[2a+(n−1)×d]

S

x−1

=S

49

−S

x

⇒

2

(x−1)

[2(1)+(x−1−1)1]=

2

49

[2+48]−

2

x

[2(1)+(x−1)1]

⇒

2

x−1

[2+(x−2)]=

2

49

[50]−

2

x

[2+x−1]

⇒

2

x−1

[x]=

2

49

[50]−

2

x

[x+1]

⇒

2

x

[x−1+x+1]=

2

49

[50]

⇒

2

x

[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.

Answer 2

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Row houses are numbers from 1,2,3,4,5…….49.

Thus we can see the houses numbered in a row are in the form of AP.

So,

First term, a = 1

Common difference, d=1

Let us say the number of xth houses can be represented as;

Sum of preceding the numbers of x = sum of following numbers of x

i.e. Sum of ( 1,2,3,….x-1) = sum of [(x+1), (x+2) ,….48,49]

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

$\tt => (x-1)/2[1+x-1] = (49-x)/2[x+1+49]$

$\tt => (x-1)x=(49-x)(x+50)$

$\tt => x²-x=49x+2450-x²-50x$

$\tt => x²-x =2450-x²-x$

$\tt => 2x²=2450$

$\tt => x²=1225$

$\tt x=√1225$

$\tt x = 35$

Therefore, the value of x is 35

Hope it's Helpful.....:)