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Arithmetic Progressions

Question
CBSEENMA10009609

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 the sum of the numbers of houses following X.

Solution

Let there be a value of x such that the sum of the numbers of the houses preceding the house numbered x is equal to the sum of the numbers of the houses following it.
That is , 1+2+3+......+(x-1) =(x-1)+(x+2)+.....+49
therefore,
1+2+3+.....+(x-1)
=[1+2+.....+x+(x+1)+.....+49]-(1+2+3+....+x)
therefore space fraction numerator straight x minus 1 over denominator 2 end fraction left square bracket 1 plus straight x minus 1 right square bracket space equals 49 over 2 left square bracket 1 plus 49 right square bracket minus straight x over 2 left square bracket 1 plus 2 right square bracket
therefore straight x left parenthesis straight x minus 1 right parenthesis equals 49 space straight x 50 minus straight x left parenthesis 1 plus straight x right parenthesis
therefore straight x left parenthesis straight x minus 1 right parenthesis plus 1 left parenthesis 1 plus straight x right parenthesis equals 49 straight x 50
straight x squared minus straight x plus straight x plus straight x squared space equals space 49 space straight x 50
therefore space straight x squared space equals 49 space straight x 25
solving space saqure
straight x space equals space 7 space straight x 5 space equals 35
Since x is not a fraction, the value of x satisfying the given condition exists and is equal to 35.