Question

the houses in a row numbered consecutively from 1 to 49. Show that there exists a value of X such that sum of no. of houses proceeding the house numbered X is equal to sum of the no. of houses following X

Answer 1

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Let us assume that the required house no. is x.
i:e Sum of the numbers of houses preceding x is equal to sum of the numbers of houses following it.
Sum of first n natural numbers is given by n*(n+1)/2.
Sum of numbers of houses preceeding x is sum of first x-1 numbers=(x-1)*x/2.
Now, Sum of numbers of houses following it is equal to sum of first n houses minus sum of first x houses = n*(n+1)/2 - x*(x+1)/2.
A/Q, They both are equal.

So, We have
(x-1)*x/2 = n*(n+1)/2 - x*(x+1)/2
=> (x-1)*x/2+(x+1)*x/2 = n*(n+1)/2
=>x*(x-1+x+1)/2 = n*(n+1)/2
=>x*2x/2 = n*(n+1)/2

Putting n=49, we get
=>x^2=49*50/2
=>x^2=49*25
=>x^2=7^2*5^2
=>x= sqrt(7^2*5^2)
Therefore,
                   x=7*5=35

Hence 35 is the number of the house such that Sum of the numbers of houses preceding it is equal to sum of the numbers of houses following it.


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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.....:)