Question

Найти четыре числа, из которых три первых образуют арифметическую прогрессию, а три последних - геометрическую. Сумма крайних чисел равна 40, а сумма средних равна 20. Помогите пожалуйста.

Answer 1

E1, e2, e3, e4

Система по условию:
a=e2-e1=(e3-e1)/2
q=e3/e2=sqrt(e4/e2)
e1+e4=40
e2+e3=20


е1=40-е4
е2=20-е3
е2-е1=е4-е3-20
е3-е1=е3+е4-40
е4-е3-20=(е3+е4-40)/2
е4-е3-20=е4/2+е3/2-20
е4-е3=е4/2+е3/2
е4/2=(е3)*3/2
е4=3*е3

е3/е2=sqrt(e4/e2)
e3/e2=sqrt(3*e3/e2)
(e3/e2)^2=3*e3/e2
e3/e2=3
e3=3*e2

q=3
e2+e3=20
e2+3*e2=20
e2=5
e3=15
e4=45
e1=40-45=-5
a=10

-5, 5, 15, 45

Answer 2

Finding the Numbers

To find four numbers where the first three form an arithmetic progression and the last three form a geometric progression, we can use the given information.

Let's denote the four numbers as a, b, c, and d.

We know that the sum of the outer numbers (a and d) is 40, and the sum of the middle numbers (b and c) is 20.

Using this information, we can set up the following equations:

Equation 1: a + d = 40 Equation 2: b + c = 20

To find the values of a, b, c, and d, we need to solve these equations simultaneously.

Solving the Equations

To solve the equations, we can use substitution or elimination. Let's use substitution in this case.

From Equation 1, we can express a in terms of d: a = 40 - d

Substituting this value of a into Equation 2, we get: (40 - d) + c = 20

Simplifying the equation, we have: c = d - 20

Now we have expressions for a and c in terms of d.

Arithmetic Progression

We also know that the first three numbers (a, b, and c) form an arithmetic progression. This means that the difference between consecutive terms is constant.

Let's denote the common difference as r.

Using this information, we can set up the following equation: b = a + r

Substituting the expressions for a and c in terms of d, we have: b = (40 - d) + r

Geometric Progression

Similarly, we know that the last three numbers (b, c, and d) form a geometric progression. This means that the ratio between consecutive terms is constant.

Let's denote the common ratio as q.

Using this information, we can set up the following equation: c = b * q

Substituting the expression for b in terms of d and r, we have: (40 - d) + r = ((40 - d) + r) * q

Solving for the Numbers

Now we have two equations involving d, r, and q:

Equation 3: b = (40 - d) + r Equation 4: (40 - d) + r = ((40 - d) + r) * q

To solve for the values of d, r, and q, we need to solve these equations simultaneously.

Let's substitute Equation 3 into Equation 4:

((40 - d) + r) = (((40 - d) + r) * q) * q

Simplifying the equation, we have: 1 = q^2

Taking the square root of both sides, we get: q = 1

Substituting this value of q back into Equation 3, we have: b = (40 - d) + r

Since q = 1, the last three numbers (b, c, and d) form an arithmetic progression.

Now we have three equations involving d, r, and q:

Equation 5: a + d = 40 Equation 6: b + c = 20 Equation 7: b = (40 - d) + r

To find the values of d, r, and q, we need one more equation. Let's use the fact that the sum of the middle numbers (b and c) is 20:

b + c = 20

Substituting the expressions for b and c in terms of d and r, we have: ((40 - d) + r) + ((40 - d) + r) * q = 20

Simplifying the equation, we have: 2(40 - d) + 2r = 20

Simplifying further, we get: **80