Many a times in arithmetic we face problems related to finding consecutive integers the sum of which is given to us. The said integers may be odd or even or neither. In this article we’ll try to understand how to solve such type of problems.
Formula for sum of consecutive integers:
As such there is no specific formula for finding sum of consecutive integers. Such problems are solved using algebra and arithmetic approach. The methods may however, vary from problem to problem. Let us look at the following example to understand this better.
Example 1: Three consecutive integers are there whose sum is 66. Find consecutive integers.
Solution: Suppose we assume that the middle or the second integer is n, then the first integer would be n-1 and the third integer would be n+1, because the integers are said to be consecutive, and we know that consecutive integers differ by 1. The three integers are therefore now: n-1, n and n+1. The sum of these integers is given to be 66. Therefore,
n-1 + n + n+1 = 66
3n = 66
n = 66/3
n = 22.
Since n was our middle integer, the other two would be n-1 = 21 and n+1 = 23. So the three consecutive integers are 21, 22 and 23.
Odd consecutive integers:
Odd consecutive integers are such that each of the integer in the list is an odd number. It may be negative or positive, but has to be an odd number. For example, …. -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9…These are odd consecutive integers. They go up to infinity in both directions.
Example 2: The sum of three consecutive odd integers is -39. Find the integers.
Solution: Here again, assume that the middle integer is n. Then the odd integer before n would be n-2 and the odd integer after n would be n+2. So the sum
= n-2 + n + n+2
= 3n = -39
n = -39/3
n = -13.
Therefore the other two odd integers would be -15 and -11. So solution: -15, -13, -11 are the required three odd integers.
Even consecutive integers:
The method of solving problems of even consecutive integers is exactly same as that for odd integers.
Example 3: Find the three even integers whose sum is 0.
Solution: If the middle integer is n, then the other two would be n-2 and n+2. Therefore the sum
= n-2 + n + n+2
= 3n = 0
n = 0/3
n = 0.
So the integers would be -2, 0, and 2
Formula for sum of consecutive integers:
As such there is no specific formula for finding sum of consecutive integers. Such problems are solved using algebra and arithmetic approach. The methods may however, vary from problem to problem. Let us look at the following example to understand this better.
Example 1: Three consecutive integers are there whose sum is 66. Find consecutive integers.
Solution: Suppose we assume that the middle or the second integer is n, then the first integer would be n-1 and the third integer would be n+1, because the integers are said to be consecutive, and we know that consecutive integers differ by 1. The three integers are therefore now: n-1, n and n+1. The sum of these integers is given to be 66. Therefore,
n-1 + n + n+1 = 66
3n = 66
n = 66/3
n = 22.
Since n was our middle integer, the other two would be n-1 = 21 and n+1 = 23. So the three consecutive integers are 21, 22 and 23.
Odd consecutive integers:
Odd consecutive integers are such that each of the integer in the list is an odd number. It may be negative or positive, but has to be an odd number. For example, …. -11, -9, -7, -5, -3, -1, 1, 3, 5, 7, 9…These are odd consecutive integers. They go up to infinity in both directions.
Example 2: The sum of three consecutive odd integers is -39. Find the integers.
Solution: Here again, assume that the middle integer is n. Then the odd integer before n would be n-2 and the odd integer after n would be n+2. So the sum
= n-2 + n + n+2
= 3n = -39
n = -39/3
n = -13.
Therefore the other two odd integers would be -15 and -11. So solution: -15, -13, -11 are the required three odd integers.
Even consecutive integers:
The method of solving problems of even consecutive integers is exactly same as that for odd integers.
Example 3: Find the three even integers whose sum is 0.
Solution: If the middle integer is n, then the other two would be n-2 and n+2. Therefore the sum
= n-2 + n + n+2
= 3n = 0
n = 0/3
n = 0.
So the integers would be -2, 0, and 2
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