Question

Three consecutive even positive integers sum of whose squares is 365

Answer 1

Answer:

let the consecutive numbers are

x - 1 , x. and x +1

ATQ

we get

(x-1)² + x² +(x+1)² =. 365

x² + 1 -2x +x² +x² +1 +2x = 365

3x² +2 = 365

x² = 363/3

x² = 121

x = 11

required numbers

are x -1 = 11-1=10

x = 11

and x +1 = 11+1 = 12

Three consecutive even positive integers sum of whose squares is 365 are 10, 11, and 12