Question : The sum of squares of 3 consecutive positive numbers is 365. The sum of the numbers is:
Option 1: 30
Option 2: 33
Option 3: 36
Option 4: 45
Answer (1)
11th Jan, 2024
Correct Answer: 33
Solution :
Let the numbers be $n–1, n$ and $n+1$.
$(n–1)^{2} + (n)^{2} + (n+1)^{2} = 365$
or, $n^2 – 2n+1+n^2+n^2+2n+1 = 365$
or, $3n^2 = 365–2$
or, $3n^2 = 363$
or, $n^2 = 121$
or, $n = 11$
So, 10, 11 and 12 are the numbers.
Their sum is (10 + 11 + 12) = 33
Hence, the correct answer is 33.
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