It is known that the sum of the four numbers in the arithmetic sequence is 40, and the product of the second number and the third number is 36 OK, I'll add points immediately

It is known that the sum of the four numbers in the arithmetic sequence is 40, and the product of the second number and the third number is 36 OK, I'll add points immediately

(a1+a2+a3+a4)=(4a1+6d)=40
A1 = (20-3d) / 2
a2*a3=(a1+d)*(a1+2d)=36
Take A1 into the second equation and calculate d, then calculate A1
I don't think so

We know that four numbers form an arithmetic sequence, the sum of them is 36, and the product of the two terms in the middle is 32

36=4(a1+a4)/2
a1+a4=18
So A2 + a3 = 18
From A2 * A3 = 32
So A3 = 2 or 16, A2 = 16 or 2
So the tolerance is 14
So A1 = 30 or - 12
A4 = - 12 or 30
A1 = 30, A2 = 16, A3 = 2, A4 = - 12 or A1 = - 12, A2 = 2, A3 = 16, A4 = 30

The sum of the three numbers is 30, and the product of the first term and the last term is 36. Find the three numbers, and the sum of the three numbers is%

Number in the middle = 30 / 3 = 10
Let the first be 10-d and the third be 10 + D
(10-d)(10+d)=36
100-d^2=36
d^2=64
D = 8 or - 8
The three numbers are: 2, 10, 18

There are four numbers, among which the first three numbers form an equal ratio sequence, the product of which is 216, and the last three numbers form an equal difference sequence, the sum of which is 36

Let these four numbers be AQ, a, AQ, 2aq − a, then AQ ⋅ a · AQ = 216a + AQ + (2aq − a) = 36. ① ② from ①, A3 = 216, a = 6 & nbsp; & nbsp; & nbsp; & nbsp; ③, ③ is substituted into ②, and 3aq = 36, q = 2. These four numbers are 3, 6, 12, 18