Question

The sum of 3 numbers in an AS is 12 and the sum of their squares is 66. find the numbers.​

Answer 1

Explanation:

Let,the three numbers be a-d,a,a+d.then,the AS

becomes a-d,a,a+d.

Acc.to question,

a-d+a+a+d=12

or,3a=12

:.a=4

also,

(a-d)²+a²+(a+d)²=66

or,a²-2ad+d²+a²+a²+2ad+d²=66

or,3a²+2d²=66

or,3×(4)²+2d²=66

or,48+2d²=66

or,2d²=18

or,d²=9

:.d=3

substituting value of a and d,we get

a-d=1

a=4

a+d=7

Hence, 1,4 and 7 are req. numbers.

Answer 2

The 3 numbers are 0, 6, and 6.

Explanation:

Let the 3 numbers be a, b, and c. We can write the following system of equations to represent the given information:

a + b + c = 12

a^2 + b^2 + c^2 = 66

Subtracting the first equation from the square of the second equation, we get:

a^2 + b^2 + c^2 - (a + b + c)^2 = 0

Expanding the square on the right side gives:

a^2 + b^2 + c^2 - a^2 - 2ab - 2ac - 2bc - b^2 - c^2 = 0

Combining like terms and simplifying the equation gives:

-4ab - 4ac - 4bc = 0

This equation tells us that the sum of the products of any two of the numbers is equal to 0. Since the sum of the products of any two of the numbers is 0, it follows that at least one of the numbers must be 0. Without loss of generality, let a=0.

Substituting a=0 into the first equation gives:

b + c = 12

Therefore, the two non-zero numbers are b and c, and their sum is 12. The only possible solution for b and c is 6 and 6, respectively. Therefore, the 3 numbers are 0, 6, and 6.