the sum of eight consecutive positive integers is 404. What is the sum of the least and greatest numbers in this sequence?

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A sequence of Integers where each integer is one more than the previous integer in the sequence are called Consecutive Integers.

If we denote the first number in the sequence by n, then

`n + (n+1) + (n+2) + (n+3) + (n+4) + (n+5) + (n+6) + (n+7) = 404`

`rArr 8n+28=404`

`rArr n=47`

Hence, the least number, n=47

The greatest number=(n+7)=47+7=54

**Therefore, the sum of the least and greatest numbers in this sequence=47+54=101.**

**Sources:**

The sum of 8 consectutive numbers is 404. Find the sum of teh first and last number in the sequence:

Since teh numbers are consecutive they form an arithmetic sequence. The terms of an arithmetic sequence can be described as

`a_n=a_1+(n-1)d` where `a_n` is the nth term,`a_1` the first term, n the number of terms, and d the common difference.

In this sequence we have n=8 and d=1.

The sum of the first n terms (n finite) of an arithmetic sequence is

`S_n=n((a_1+a_n)/2)`

Here the sum is given as 404 and we know n=8 so:

`404=8((a_1+a_n)/2)`

`404=4(a_1+a_n)`

`a_1+a_n=101`

**The sum of the first and last numbers of the sequence is 101.**

Another way to see it is that since the numbers are consecutive, the first and last will be equally far away from the average (one to the left of the average and one to the right of it on the number line). Therefore, they add up to twice the average. For example, in the list 3,4,5,6, the sum of the first and last numbers is 9, which is twice the average of the list, which is 4.5.

In our case, the average of our numbers is `404/8=50.5.` **Thus the first and last numbers must add to twice this, or 101.**

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