Three consecutive natural numbers are such that the square of the middle number exceeds the difference of the squares of the other two by $60$ .

Open in App

Solution

Let three consequenture natural numbers are n-1, n, and n+1
Now
Given $(n)^{2} = (n + 1)^{2} - (n - 1)^{2} + 60$
$n^{2} = n^{2} + 1 + 2 n - n^{2} - 1 + 2 n + 60$
$n^{2} - 4 n - 60 = 0$
$n^{2} - 10 n + 6 n - 60 = 0$
$n (n - 10) + 6 (n - 10) = 0$
$(n + 6) (n - 10) = 0$
[n=-6] and [n=10]
Hence three numbers are -7,-6,-5 or 9,10 and 11
But numbers $\neq - 7, - 6, - 5$ then three numbers are 9,10 & 11

Suggest Corrections

Similar questions

Three consecutive natural numbers are such that the square of the middle number excceds the difference of the squares of the other two by 60.

Assume the middle number to be x and form a quadratic equation satisfying's the above statement. Hence; find the three numbers.

1

3, 5

7

The sum of the squares of three consecutive natural numbers is 2030.What is the middle number?

Join BYJU'S Learning Program