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If, for a positive integer n, the quadratic equation, x(x+1)+(x+1)(x+2)+.... +( n 1) x + − (x+n)=10n has two consecutive integral solutions, then n is equal to : 

Asked by kmondaldgms 23rd August 2018, 11:03 PM
Answered by Expert
Answer:
begin mathsize 18px style sum from straight r equals 1 to straight n of open parentheses straight x plus straight r minus 1 close parentheses open parentheses straight x plus straight r close parentheses equals 10 straight n
sum from straight r equals 1 to straight n of open parentheses straight x squared plus open parentheses 2 straight r minus 1 close parentheses straight x plus open parentheses straight r squared minus straight r close parentheses close parentheses equals 10 straight n
solving
straight x squared plus nx plus open parentheses fraction numerator straight n squared minus 31 over denominator 3 end fraction close parentheses equals 0
difference space of space roots space equals space 1
open parentheses straight alpha minus straight beta close parentheses squared equals 1
open parentheses straight alpha plus straight beta close parentheses squared minus 4 αβ equals 1
straight n squared over 1 minus 4 open parentheses fraction numerator straight n squared minus 31 over denominator 3 end fraction close parentheses equals 1
3 straight n squared minus 4 straight n squared plus 124 equals 3
straight n squared equals 121
straight n equals 11
end style
Answered by Expert 27th August 2018, 12:08 PM
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