Question

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

Answer 1

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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

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