Asked by Vic
The sum of 25 consecutive integers is 1000. What is the smallest integer used as an addend.
Answers
Answered by
Steve
the sum of the first n integers is
n(n+1)/2
Now, we have 25 integers starting somewhere else. That is, if the first one is k+1, then we add k to each value, giving us
nk + n(n+1)/2
So, using n=25,
25k + 25*26/2 = 1000
k + 13 = 40
k = 27
The numbers are 28,...,52
Checking our work, recall that the sum of n terms of an arithmetic sequence is
n(Ao+An)/2
= 25(28+52)/2 = 25*80/2 = 1000
n(n+1)/2
Now, we have 25 integers starting somewhere else. That is, if the first one is k+1, then we add k to each value, giving us
nk + n(n+1)/2
So, using n=25,
25k + 25*26/2 = 1000
k + 13 = 40
k = 27
The numbers are 28,...,52
Checking our work, recall that the sum of n terms of an arithmetic sequence is
n(Ao+An)/2
= 25(28+52)/2 = 25*80/2 = 1000
Answered by
Not good
Holy crap you know math
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.