找出和为 2 的四个连续整数
代数表达式是由变量和分量组合形成的项的组合。代数表达式中的项通过使用加法或减法的数学运算连接在一起。常量可以是整数。项的数量定义了代数表达式。代数表达式中可能有一个或多个变量。
连续数字
连续数字是在数轴上同时出现的数字。所有这些数字之间都有 +1 或 -1 的差异。一些连续的自然数是1、2、3、4、5……在连续数的情况下，任何前继对之间的差异都是固定的。
找出和为 2 的四个连续整数
解决方案：
Approach 1: Using the AP formula
Lets assume the numbers to x, x + 1, x + 2, and x + 3 respectively.
If clearly notice, the numbers form an AP:
Where the first term, a = x
Common difference, d = 1
Total number of terms, n = 4
Sum of n terms in an A.P = n/2[2a + (n – 1)d]
On substituting the values,
2 = 4/2[2 × x + (4 – 1) × 1]
2 = 2[2x + 3]
1 = 2x + 3
On solving,
2x = -2
x = -1
Therefore, the first number is -1.
The other numbers as,
x + 1 = 0, x + 2 = 1, x + 3 = 2.
Therefore, the numbers are -1, 0, 1, 2. The sum of these numbers is 2.
Approach 2: Using the variable method
Lets assume the starting number to be x.
Therefore, identifying the numbers,
x, x + 1, x + 2 and x + 3.
Now,
Summation of these numbers is equivalent to 2.
x + x + 1 + x + 2 + x + 3 = 2
On solving the equation,
4x + 6 = 2
Equating the values,
4x = -4
x = -1
Therefore, the value of x is equivalent to -1.
Now, computing the numbers,
x = -1
x + 1 = 0
x + 2 = 1
x + 3 = 2
Therefore, the numbers are -1, 0, 1, 2. The sum of these numbers is 2.
示例问题
问题 1：找出总和为 657 的三个连续整数。
解决方案：
Assume the first number be x
Then the next two numbers will be x + 1, x + 2
Thus,
x + x +1 + x + 2 = 657
3x + 3 = 657
3x = 657 – 3
3x = 654
x = 654/3
x = 218
Therefore the integers are
x = 218
x + 1 = 219
x + 2 = 220
218 + 219 + 220 = 657
问题2：如果三角形边长的度量是连续的偶数。然后找出周长为 90 厘米的三角形的所有长度？
解决方案：
Here as the sides are the consecutive even numbers, so add 2 to the previous side
Thus,
Length of side 1 = x
Length of side 2 = x + 2
Length of side 3 = x + 4
The perimeter of the triangle is sum of all the sides
Therefore,
Perimeter of the triangle = x + x + 2 + x + 4
90 = 3x + 6
3x = 90 – 6
x = 84/3
x = 28
Therefore,
Length of side 1 = x = 28
Length of side 2 = x + 2 = 30
Length of side 3 = x + 4 = 32
问题3：如果三个连续整数之和为-180。找到最大的整数？
解决方案：
Assume the first integer be x
Thus,
The three integers will be
x, x + 1, x + 2
x + x + 1 + x + 2 = -180
3x + 3 = -180
3x = -180 – 3
3x = -183
x = -183/3
x = -61
x + 1 = -61 + 1 = -60
x + 2 = -61 + 2 = -59
Thus, the largest integer is -59
Therefore,
-59 + (-60) + (-61) = -180
问题4：如果连续五个整数之和为200，求第二个整数？
解决方案：
Assume the smallest integer be x
Therefore sum of all the integers will be
x + x + 1 + x + 2 + x + 3 + x + 4 = 200
5x + 10 = 200
5x = 200 – 10
5x = 190
x = 190/5
x = 38
Therefore,
The first integer will be 38
The second integer will be 38 + 1 = 39
The third integer will be 38 + 2 = 40
The forth integer will be 38 + 3 = 41
The fifth integer will be 38 + 4 = 42
Thus,
38 + 39 + 40 + 41 + 42 = 200
The second integer is 39.
问题 5：找出数字 -23 后面的五个连续奇数。
解决方案：
Here, to find 5 consecutive odd integers following -23
As it is known that of consecutive odd integers differ by 2
Thus, the integers will be like
x, x + 2, x + 4, x + 6, x + 8
If assumed x = -23
Then,
The first integer will be x = -23
The second integer will be x + 2 = -23 + 2 = -21
The third integer will be x + 4 = -23 + 4 = -19
The forth integer will be x + 6 = -23 + 6 = -17
The fifth integer will be x + 8 = -23 + 8 = -15.