Question 1093785: Part (a): Find the sum of a + (a + 1) + (a + 2) + ... + (a + n - 1)in terms of a and n.
Part (b): Find all pairs of positive integers (a,n) such that n>=2 and a + (a + 1) + (a + 2) + ... + (a + n - 1) = 100.
Found 2 solutions by ikleyn, greenestamps:
Answer by ikleyn(52788)   (Show Source):
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
I have always avoided the standard textbook form of the formula for the sum of n terms of an arithmetic series with first term a and common difference d=1:
I find this formula awkward to use; I always use a less formal formula. HOWEVER, this formula works well to solve the second part of your question. In that part of your question, we want to find all the pairs of positive integers (a,n) for which the sum of the n consecutive integers starting with a is equal to 100. So
Now 2a, n, and 1 are integers, so 2a+n-1 is an integer; that means 200/n must be an integer. So we can find all the solutions to the problem by trying all the values of n (greater than or equal to 2, according to the problem) that make 200/n an integer, and finding which ones have an integer a that makes 2a+n-1 equal to that same integer.
If n=2, then
and we would need to have
but the value of a for the equation is not an integer.
If n=4, then
and we would need to have
and again the value of a for the equation is not an integer.
If n=5, then
and we would need to have
This equation DOES have an integer value for a: a=18.
If n=8, then
and we would need to have
This equation also has an integer value for a: a=9.
It turns out that no other values of n that are factors of 200 give integer values for a.
So we have two solutions for the second part of your problem: (a,n) = (18,5) and (a,n) = (9,8).
It is easy to check that these two solutions give sums of 100:
(18,5) --> 5 consecutive integers starting with 18: 18+19+20+21+22 = 100
(8,8) --> 8 consecutive integers starting with 9: 9+10+11+12+13+14+15+16 = 100
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