The sum of squre three consecutive numbers is 365.Find out smallest number?

• Let x-1,x,x+1 are three consecutive numbers
  then according to the problem,
  (x-1)^2 + x^2 + (x+1)^2 =365
  3x^2=365-2=363
  x^2=121
  x=-11,11
  hence, smallest number = x-1=-12 or 10 ans.
• 11 years agoHelpfull: Yes(4) No(2)
• let three cosecutive numbers are : (x-1),x,(x+1)
  So, (x-1)^2+x^2+(x+1)^2=365
  => 3x^2+2=365
  => x^2=(365-2)/3
  x = 11 or -11
  therefore, the numbers are: (10, 11 , 12) or (-12, -11, -10)
  So the smallest number is: -12
  Ans: -12
  
• 11 years agoHelpfull: Yes(1) No(1)
• let three cosecutive numbers are : (x-1),x,(x+1)
  now,
  (x-1)^2+x^2+(x+1)^2=365
  x^2-2x+1+x^2+x^2+2x+1=365
  3x^2+2=365
  x^2=(365-2)/3
  x=11
  therefore, the numbers are: 10, 11 , 12
  smallest number is: 10 ans
  
• 11 years agoHelpfull: Yes(0) No(1)
• Three consecutive numbers are a-1,a,a+1.
  Sum of the squares of 3 consecutive number is (a-1)^2+a^2+(a+1)^2 = 365
  a^2-2a+1+a^2+a^2+2a+1 = 365
  3a^2+2 = 365
  3a^2 = 363
  a^2 = 121
  a = +11 or -11
  The consecutive numbers are,-12,-11,-10 & 10,11,12
  the smallest number is -12.
• 11 years agoHelpfull: Yes(0) No(0)
• ans=10,-12
  (a-1)^2+(a)^2+(a+1)^2=365
  3a^2+2=365
  3a^2=363
  a^2=121
  a=11,-11
  smallest no.=(a-1)=(11-1)=10.
  OR (-11-1)= -12.
• 11 years agoHelpfull: Yes(0) No(0)