6, 7, 8
Detailed Answer:
Assume the numbers are: n-1, n, n+1 ...... (1)
then, (n-1)2 + n2 + (n+1)2 = 149
3n2 + 2 = 149
3n2 = 147
n2 = 49
n = 7 (or -7 but not valid)
substitute in eq 1
the answer is as above 6, 7, 8
The numbers are 12 and 14.
The numbers are 8 and 9.
6,7,8 those numbers squared are 36,49,64 which add to 149
Consecutive numbers are whole numbers whose difference is 1.
Let the two consecutive numbers be ( n ) and ( n + 1 ). The difference of their squares can be expressed as ( (n + 1)^2 - n^2 ), which simplifies to ( 2n + 1 ). Setting this equal to 25, we get the equation ( 2n + 1 = 25 ). Solving for ( n ), we find ( n = 12 ), so the two consecutive numbers are 12 and 13.
The numbers are 12 and 14.
The numbers are 12 and 14.
8
The numbers are 13 and 14.
62 and 63
15 and 16
17 and 18
The numbers are 8 and 9.
6,7,8 those numbers squared are 36,49,64 which add to 149
The difference between the squares of two consecutive integers j and j+1 is |2j+1|. There are therefore two such pairs where this quantity is 17:-9 and -88 and 9
Consecutive numbers are whole numbers whose difference is 1.
24 and 25, which are (49-1)/2 and (49+1)/2