4. сумма четвертых и пятых членов арифметической прогрессии равна 14, а разница равна 4 найдите арифметическую прогрессию

Given that the sum of the fourth and fifth terms of an arithmetic progression is 14, and the difference is 4, we can solve for the arithmetic progression.

Let the first term be "a" and the common difference be "d".

The fourth term can be expressed as "a + 3d", and the fifth term as "a + 4d".

According to the problem, (a + 3d) + (a + 4d) = 14.

This simplifies to 2a + 7d = 14.

Since the difference is given as 4, we have d = 4.

By substituting the value of d into the equation 2a + 7d = 14, we find that 2a + 7(4) = 14.

Simplifying further, we have 2a + 28 = 14.

Subtracting 28 from both sides gives 2a = -14.

Finally, dividing by 2 gives a = -7.

Therefore, the arithmetic progression is: -7, -3, 1, 5, 9, ...