+0

Arithmetic Sequences Question 2

0
395
7

Find all pairs of positive integers (a, n) such that n ≥ 2 and

a + (a+1) + (a+2) + ... + (a+n-1) = 100.

Jan 17, 2019

#1
0

here is hint: when n is even, you can pair up numbers, when n is odd, find the middle term, because it will be the average.

HOPE THIS HELPED!

Jan 17, 2019
edited by asdf335  Jan 17, 2019
#2
+1

18 + 19 + 20 + 21 + 22 =

18 + 22 + 19 + 21 + 20 =

40 + 40 + 20 =

100

So   ( a, n)  =  (18, 5)   Jan 17, 2019
#3
+2

How did you work that out Chris?

Melody  Jan 17, 2019
#4
0

Also, the following: 9 + 10 + 11 + 12 + 13 + 14 + 15 +  16 = 100

Jan 17, 2019
#5
+2

Are you both doing it by trial and error?  The question does say ALL

Jan 18, 2019
edited by Melody  Jan 18, 2019
#6
+3

We can write:

a + (a+1) + (a+2) + ... + (a+n-1) = na + (n-1)n/2

Hence we have:  na + (n-1)n/2 = 100, which we can rewite as:  n2 + (2a-1)n - 200 = 0

Solve this for n in terms of a: $$n=\frac{\sqrt{4a^2-4a+801}-2a+1}{2}$$  (the other solution results in negative values for n).

Let a run from 1 to 100 to find the only integer solutions are:  (a,n) = (9,8) and (18,5) as found by Chris and Guest.

(There is also a = 100 and n = 1, but the question requires n>=2).

(Alternatively, solve for a in terms of n and run through n = 2 to 100, and look for integer solutions for a)

Alan  Jan 18, 2019
edited by Alan  Jan 18, 2019
#7
+2

Thanks Alan :)

I did that, I just didn't think to finish it with a run.

I was looking for some other way to do it.

But I guess I could do it in EXCEL easily enough. :/

Melody  Jan 18, 2019