Question

The houses in a row are numbered consecutively from 1 to 49.Show that there is a value of 'x' such that the sum of numbers 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'.​

Answer 1

Answer: x=34

You can do this by trial and error or using the long unnecessary method of arithmetic progression

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.....:)