The four numbers in turn form an arithmetic sequence, and the sum of squares of the four numbers is 94. The product of the first and the last two numbers is 18 less than the product of the middle two numbers

The four numbers in turn form an arithmetic sequence, and the sum of squares of the four numbers is 94. The product of the first and the last two numbers is 18 less than the product of the middle two numbers

Let four numbers be A-D, a, a + D, a + 2D, so (A-D) ^ 2 + A ^ 2 + (a + D) ^ 2 + (a + 2D) ^ 2 = 944a ^ 2 + 4AD + 6D ^ 2 = 94 (A-D) (a + 2D) + 18 = a (a + D) a ^ 2 + ad-2d ^ 2 + 18 = a ^ 2 + add ^ 2 = 9D = 3D = - 34a ^ 2 + 4AD + 6D ^ 2 = 94 If d = 3, a ^ 2 + 3a-10 = 0A = - 5, a = 2 if d = - 3, a ^ 2-3a-10 = 0A = 5, a = - 2, so

The sum of squares of the arithmetic sequence composed of four numbers is 94, and the product of the first and the last two numbers is 18 less than the product of the middle two numbers

Hello
Let's set these four numbers as
A-d A A+d A+2d
And because the sum of squares is 94
The product of the first and the last two numbers is 18 times smaller than that of the middle two numbers
We get (A-D) ^ 2 + A ^ 2 + (a + D) ^ 2 + (a + 2D) ^ 2 = 94
A(A+d)=18+(A-d)(A+2d)
Just solve a and D
thank you

The arithmetic sequence with the first term of 24 is negative from the 10th term

a9=24-8d≥0
a10=24-9d