Question

Сума трьох чисел, які є послідовними членами арифметичної прогресії, дорівнює 21. Якщо друге число зменшити на 1, а третє збільшити на 1, то одержимо три послідовних члени геометричної прогресії. Знайдіть ці числа.

Answer 1

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Answer 2

Problem Analysis

We are given that the sum of three consecutive terms of an arithmetic progression is 21. If the second term is decreased by 1 and the third term is increased by 1, we obtain three consecutive terms of a geometric progression. We need to find these three numbers.

Solution

Let's assume that the three consecutive terms of the arithmetic progression are a-d, a, and a+d, where 'a' is the second term and 'd' is the common difference.

According to the given information, the sum of these three terms is 21: (a-d) + a + (a+d) = 21

Simplifying the equation, we get: 3a = 21 a = 7

Now, if we decrease the second term by 1 and increase the third term by 1, we obtain three consecutive terms of a geometric progression. Let's call these terms b, c, and d, where 'b' is the first term, 'c' is the second term, and 'd' is the third term.

The terms b, c, and d form a geometric progression, so we have: c/b = d/c

Substituting the values of b, c, and d, we get: (a-1)/a = (a+1)/(a-1)

Substituting the value of a as 7, we can solve this equation to find the values of b, c, and d.

Calculation

Let's solve the equation (a-1)/a = (a+1)/(a-1) for a = 7:

(7-1)/7 = (7+1)/(7-1) 6/7 = 8/6 36 = 56

The equation is not true, which means there is no solution that satisfies the given conditions.

Conclusion

Based on the given information, there are no three consecutive terms of an arithmetic progression that satisfy the condition of having their sum equal to 21 and also form three consecutive terms of a geometric progression when the second term is decreased by 1 and the third term is increased by 1.