# Three consecutive natural numbers ae such that the square of the middle number exceeds the difference of the squares of the other two numbers by 60

2
by vasistsanjana

2016-01-02T16:06:22+05:30
Since the middle number of the three consecutive numbers is x, the other two numbers are x -1 and x +1.
According to the given condition, we have
x² = [(x +1)² -(x -1)²] +60
x² = (x² +2x +1) -(x² -2x +1) +60 = 4x +60
x² -4x -60 = 0 => (x -10)(x +6) = 0
x = 10 or -6
Since x is a natural number, we get x = 10
Hence the three numbers are 9, 10, 11
2016-01-02T17:41:11+05:30
Question: Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two numbers by 60.

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There are three consecutive integers (same as, natural
numbers).

let the smallest of the numbers be
.

Values of
:

1st no. is .
2nd no. is .
3rd no. is    .

Condition: The square of the 2nd no. (middle number) is 60 more than the difference of the squares of other two numbers.

Solution:

=>

=>

=>

=>

=>

=>

=>

=>

=>

=>

Since,  is an irrational number, so we will not further simplify it.

By putting the values of ,

1st no. is .
2nd no. is .
3rd no. is .

ANSWER: The values of the three consecutive numbers:

1st no. is .
2nd no. is .
3rd no. is .

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