M4maths Previous Puzzles

27 Dec 2019 Arithmetic

The sum of squares of three successive integers is 5810. Sum of these integers is?

OPtion

1) 122
2) 124
3) 126
4) 128
5) 116
6) 132
7) 118
8) 110
9) 112
10) None of these

Solution

(a-1)^2 + a^2 + (a+1)^2 = 5810
=> a^2 -2a + 1 + a^2 + a^2 + 2a + 1 = 5810
=> 3a^2 + 2 = 5810
=> a^2 = (20174-2)/3 = 1936
=> a = 44
So the 3 numbers are 43, 44 and 45 and their sum is 132

Correct Option: 6 Best Solution (2)