wikiHow to Add 5 Consecutive Numbers Quickly
Four Methods:Using the Middle NumberUsing the Highest NumberUsing the Lowest NumberUsing Other Amounts of Consecutive Numbers
Bet someone you can add five consecutive integers faster than they can. You can use this as a bar trick, with your friends, or (if you are a student) impress your teacher!
Steps
Method 1 Using the Middle Number
 1Mentally multiply the middle number by 5 ... finished!? That's all you do!. e.g. with your example 53 X 5 = 265. Here's how to multiply this mentally:
 First separate 53 into 50 and 3.
 Now 50 X 5 = 250.
 Also 3 X 5 = 15.
 Now add those separate answers. 250 + 15 = 265.
 2Learn how it works:
 Let the smallest number be (x  2). Then the other 4 are (x  1), (x), (x + 1) and (x + 2).
 The sum: (x  2) + (x  1) + (x) + (x + 1) + (x + 2) = 5x
 Using the above method: 10x/2 = 5x
Method 2 Using the Highest Number
Method 3 Using the Lowest Number
Method 4 Using Other Amounts of Consecutive Numbers
 1To add four consecutive numbers, multiply the highest by 4 and subtract 6.
 2To add six consecutive numbers, multiply the highest by 6 and subtract 15.
 3To add seven consecutive numbers, multiply the highest by 7 and subtract 21.
 4To add eight consecutive numbers, multiply the highest by 8 and subtract 28.
Tips
 You can add any sequence of consecutive numbers, even or odd, regardless of the number of integers the sequence contains, by adding the first and the last number in the sequence, dividing it by two and multiplying the answer by the number of integers in the sequence. In algebra, we can say ((a+b)/2)*n, or rearrange it to n*(a+b)/2 to remove a set of parenthesis.
 Method 2 can be used for any odd amount of consecutive numbers, but instead of using "5x" you must use "(amount of consecutive numbers)x"
 e.g. in 6 + 7 + 8, seven is x.
 (3)7 = 21, and 6 + 7 + 8 = 21
Advanced Usage
 They don't need to be consecutive numbers. They just need to be a sequential subset of any linear equation. ( The examples above use the linear equation of x = c + 1 * n )
 For example, let's use the linear equation x = 10 + 7y, therefore, {xϵN 17,24,31,38,45,...}

 So if we use: 17, 24, 31, 38, 45
 31 x 10 = 310 and 310/2 = 155

 They also do not need to be integers. Let's use the linear equation x = 1 + y / 20, therefore, {xϵN 1.05, 1.1, 1.15, 1.2, 1.25, ...}

 So if we use: 1.05, 1.1, 1.15, 1.2, 1.25
 1.15 x 10 = 11.5 and 11.5/2 = 5.75

 They also do not need to be positive values. The set can contain negative numbers, positive numbers, or both.
 This method can be extended to be used (as above) for any ODD number of consecutive integers > 5, 7, 13, 25, 99 consecutive as long as you can locate the median digit and then multiply it be the number of integers. (Example 12, 13, 14, 15, 16, 17, 18, 19, 20 = 144 = 16(median) x 9(integer count). This can be most impressive if coupled with the simple trick of multiplying by 11.
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