Question

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

Answer 1

Hey there,

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.
So, Sum of numbers of houses preceding 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.

Hope that helps!

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.....:)