First, let's find the sum of the 250 consecutive integers starting from -125.

The sum of a series of consecutive integers can be found using the formula for the sum of an arithmetic series:
\[ S = \frac{n}{2} (a_1 + a_n) \]
where \( S \) is the sum of the series, \( n \) is the number of terms, \( a_1 \) is the first term, and \( a_n \) is the last term.

In this case, \( n = 250 \), \( a_1 = -125 \), and since the integers are consecutive, the last term \( a_n \) will be \( a_1 + (n - 1) \), which is \( -125 + (250 - 1) = -125 + 249 = 124 \).

Now, let's calculate the sum:
\[ S = \frac{250}{2} (-125 + 124) \]
\[ S = 125 (-1) \]
\[ S = -125 \]

Next, let's find the product of the first 10 even integers. The first 10 even integers are 2, 4, 6, 8, 10, 12, 14, 16, 18, and 20. The product of these integers is:
\[ P = 2 \times 4 \times 6 \times 8 \times 10 \times 12 \times 14 \times 16 \times 18 \times 20 \]

To simplify the calculation, we can pair the numbers and use the properties of multiplication:
\[ P = (2 \times 10) \times (4 \times 8) \times (6 \times 12) \times (14 \times 18) \times (16 \times 20) \]
\[ P = 20 \times 32 \times 72 \times 252 \times 320 \]

We can further simplify by breaking down the numbers into prime factors and then multiplying, but for the sake of this calculation, let's just multiply the numbers directly:
\[ P = 20 \times 32 \times 72 \times 252 \times 320 \]
\[ P = 640 \times 72 \times 252 \times 320 \]
\[ P = 46080 \times 252 \times 320 \]
\[ P = 11612160 \times 320 \]
\[ P = 3715884800 \]

Finally, we multiply the sum of the series by the product of the first 10 even integers:
\[ Total = S \times P \]
\[ Total = -125 \times 3715884800 \]
\[ Total = -464485600000 \]

So, the sum of 250 consecutive integers starting from -125 inclusive, in an increasing order, multiplied by the product of the first 10 even integers is -464485600000.
먼저 -125에서 시작하는 연속된 250개의 정수의 합을 구해 보겠습니다.

연속된 정수의 합은 산술 급수의 합 공식을 사용하여 구할 수 있습니다:
\[ S = \frac{n}{2} (a_1 + a_n) \]
여기서 \( S \)는 급수의 합, \( n \)은 항의 수, \( a_1 \)은 첫 번째 항, \( a_n \)은 마지막 항입니다.

이 경우 \( n = 250 \), \( a_1 = -125 \), 정수가 연속적이므로 마지막 항 \( a_n \)은 \( a_1 + (n - 1) \), 즉 \( -125 + (250 - 1) = -125 + 249 = 124 \).

이제 합계를 계산해 보겠습니다:
\[ S = \frac{250}{2} (-125 + 124) \]
\[ S = 125 (-1) \]
\[ S = -125 \]

다음으로 처음 10개의 짝수 정수의 곱을 구해 보겠습니다. 처음 10개의 짝수 정수는 2, 4, 6, 8, 10, 12, 14, 16, 18, 20입니다. 이 정수의 곱은 다음과 같습니다:
\[ P = 2 \times 4 \times 6 \times 8 \times 10 \times 12 \times 14 \times 16 \times 18 \times 20 \]

계산을 단순화하기 위해 숫자를 쌍으로 묶고 곱셈의 속성을 사용할 수 있습니다:
\[ P = (2 \times 10) \times (4 \times 8) \times (6 \times 12) \times (14 \times 18) \times (16 \times 20) \]
\[ P = 20 \times 32 \times 72 \times 252 \times 320 \]

수를 소인수로 분해한 다음 곱하면 더 단순화할 수 있지만, 이 계산을 위해 숫자를 직접 곱하겠습니다:
\[ P = 20 \times 32 \times 72 \times 252 \times 320 \]
\[ P = 640 \times 72 \times 252 \times 320 \]
\[ P = 46080 \times 252 \times 320 \]
\[ P = 11612160 \times 320 \]
\[ P = 3715884800 \]

마지막으로, 급수의 합에 처음 10개의 짝수 정수의 곱을 곱합니다:
\[ Total = S \times P \]
\[ Total = -125 \times 3715884800 \]
\[ Total = -464485600000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000