The sum of squares of three successive integers is 26510. Sum of these integers is?

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03 Mar 2020 Arithmetic

The sum of squares of three successive integers is 26510. Sum of these integers is?

OPtion

1) 221
2) 282
3) 223
4) 281
5) 224
6) 286
7) 288
8) 226
9) 228
10) None of these

Solution

(a-1)^2 + a^2 + (a+1)^2 = 26510
=> a^2 -2a + 1 + a^2 + a^2 + 2a + 1 = 26510
=> 3a^2 + 2 = 26510
=> a^2 = (20174-2)/3 = 8836
=> a = 94
So the 3 numbers are 93, 94 and 95 and their sum is 282

Correct Option: 2 Best Solution (3)