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Arithmetic Progressions

Question
CBSEENMA10007924

The houses of a row are numbered consecutively from 1 to 49. Show that there is 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. Find this value of x.

Solution

The consecutive numbers on the houses of a row are 1, 2, 3, ... 49.
Clearly this list of number forming an A .P.
Here, a = 1,d = 2 – 1 = 1
According to the question,

straight S subscript straight x minus 1 end subscript equals straight S subscript 49 minus straight S subscript straight x
rightwards double arrow space space space space fraction numerator straight x minus 1 over denominator 2 end fraction left square bracket 2 straight a plus left parenthesis straight x minus 1 minus 1 right parenthesis straight d right square bracket
equals 49 over 2 left square bracket 2 straight a plus left parenthesis 49 minus 1 right parenthesis straight d right square bracket minus straight x over 2 left square bracket 2 straight a plus left parenthesis straight x minus 1 right parenthesis 1 right square bracket
space space space space space space space space space space space space space space space space space space space open square brackets because space straight S subscript straight n equals straight n over 2 left square bracket 2 straight a plus left parenthesis straight n minus 1 right parenthesis straight d right square bracket close square brackets
rightwards double arrow space fraction numerator straight x minus 1 over denominator 2 end fraction left square bracket 2 left parenthesis 1 right parenthesis plus left parenthesis straight x minus 2 right parenthesis left parenthesis 1 right parenthesis right square bracket
rightwards double arrow space space 49 over 2 left square bracket 2 left parenthesis 1 right parenthesis plus left parenthesis 48 right parenthesis left parenthesis 1 right parenthesis right square bracket minus straight x over 2 left square bracket 2 left parenthesis 1 right parenthesis plus left parenthesis straight x minus 1 right parenthesis straight d right square bracket
rightwards double arrow space space space fraction numerator straight x minus 1 over denominator 2 end fraction left square bracket straight x right square bracket equals 1225 minus fraction numerator straight x left parenthesis straight x plus 1 right parenthesis over denominator 2 end fraction
rightwards double arrow space space space space fraction numerator left parenthesis straight x minus 1 right parenthesis left parenthesis straight x right parenthesis over denominator 2 end fraction plus fraction numerator straight x left parenthesis straight x plus 1 right parenthesis over denominator 2 end fraction equals 1225
rightwards double arrow space space space space space straight x over 2 left parenthesis straight x minus 1 plus straight x plus 1 right parenthesis equals 1225
rightwards double arrow space space space space space space straight x squared equals 1225
rightwards double arrow space space space space space space straight x space equals space square root of 1225
rightwards double arrow space space space space space space straight x space equals space 35
Hence the required values of x is 35.