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# KCC's Quizzes: Sum of Consecutive Integers

Let’s consider positive integer numbers n and k. The number of consecutive terms starting from n. The series is then: n, n+1, n+2, ..., n+(k‐1).

1. If n=100 and k=100, what is the sum of the terms in that series?

2. By reversing the problem: what must be the number of terms of the series starting from n=120 and with a sum=8970?

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[edited by: emassa at 7:43 PM (GMT -5) on 3 Jan 2023]
• To be more detailed: n is the first integer of the series and k is the number of following integers to be summed. In other words, the series has k terms, starts with n and ends with n+k-1. It's easy...

• 1) 19,900

2) K = 60

• 1) sum = (100 +199)*(100/2) = 14950

2)  (120 + 120+(k-1))*(k/2) = 8970 ==> k^2 + 239k -17940 =0  ==> k=60

• The sum must be the number of terms times the average of the terms.  The number of terms is k, that's the easy part.  The average of the terms is n + (k-1)/2.  So the sum will be kn + k(k-1)/2 = 0.5k^2 + k(n - 0.5).  Plugging in 100 for both n & k yields the first answer, 5000 + 9950 = 14950.
Solving for k, entailing a quadratic, with a sum S of 8970 and a starting n=120...
k = 1-2n + SQRT[(2n-1)^ +8S] / 2

• or I just choked in a rush!  :D

• As n=100 and k=100, series is 100,101,...199

Sum(k=1,100) of n+(k-1) is

k*n = (k(k+1))/2 - k

2. it is a second degree equation, as k>0 answer is k = 60

• correction in typing, k*n+(k(k+1))/2 - k

• Oops revised response.

1) 14,950

2) K = 60

• Thanks Jeffrey, I already put my comment to you!

• Thanks Brian for your prompt feedback! I have sent you my comment directly to you.