Question

houses in a row are numbered consequently from 1 to 49.show that there exist a value of X such that sum of number of houses preceding the the house X IS EQUAL TO SUM OF NUMBER OF HOUSES FOLLOWING X

Answer 1

There are houses from 1 to 49 and there is x in between
1,2,3,4........x........ 47,48,49
we have to prove
$s(x - 1) = s(49) - s(x)$
$s(x - 1) = \frac{x - 1}{2} (1 + x - 1) \\ s(x - 1) = \frac{x - 1}{2} (x) = \frac{ {x}^{2} - x }{2}$
$s(49) = \frac{49}{2} (1 + 49) \\ = \frac{49}{2} (50) \\ = 1225$
$s(x) = \frac{x}{2} (1 + x) \\ = \frac{x + {x}^{2} }{2}$
$s(x - 1) + s(x) = s(49)$
$\frac{ {x}^{2} - x }{2} + \frac{ {x}^{2} + x }{2} = 1225 \\ \frac{ {2x}^{2} }{2} = 1225 \\ {x}^{2} = 1225 \\ x = \sqrt{1225 } \\ x = 35$

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.....:)