Latest Bank Exam Aptitude Question SOLUTION: The sum of squre three consecutive numbers is 365.Find out smallest number?

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

Asked In Bank Exam jaya mishra (11 years ago)

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IQBAL MEHDI  (11 years ago)

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.

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Rakesh Kumar Kohli  (11 years ago)

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

Is this solution Helpfull? Yes (1) | No (1)  

RAHU SHARMA  (11 years ago)

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

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Balakrishnan M  (11 years ago)

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.

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Anjali Kaushik  (11 years ago)

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.

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