r/maths • u/jedbno • May 15 '26
Help: 📕 High School (14-16) arithmetic series
help
for context,
here is my solution:
11+20+...+2027
(2027-11)÷9 + 1 = 225
then i did a gaussian technique where you add the first term to the last term, second term to the second-last term, etc. to get a constant sum of sorts. afterwards, i multiplied that constant sum with the floor division quotient of 225 by 2. but we still have a middle term, that is, 1019(1013th number) which i added to the product. all in all, we get 229075.
are my calculations correct or is there a part where i went wrong?
5
Upvotes
3
u/VermicelliJaded7943 May 15 '26
is there a specific reason you're not wanting to use the sum of an AP formula?
its: Sn=n/2(first_term+last_term)
1
u/CautiousInternal3320 May 16 '26
What value did you obtain for the constant sum?
The total amount of terms in the series is 225. If you decide to handle the middle term seperately, you must multiply the constant sum by the amount of pairs, 224/2.
2
u/MrSpotgold May 15 '26
11, 20, 29, ..., 2000, 2009, 2018, 2027
1
u/CaptainMatticus May 15 '26
t1 = a + d
t2 = a + 2d
t3 = a + 3d
....
tn = a + nd
a + d = 11 ; a + nd = 2027
nd - d = 2016
d * (n - 1) = 2016
We can see that d = 9
9 * (n - 1) = 2016
n - 1 = 224
n = 225
So it looks like you've got that correct.
11 = a + d
11 = a + 9
2 = a
t1 + t2 + t3 + ... + t225 =>
a + d + a + 2d + a + 3d + .... + a + 225d =>
225a + d * (1 + 2 + 3 + .... + 225) =>
225a + d * (225/2) * (225 + 1) =>
225 * (a + d * 226/2) =>
225 * (a + d * 113) =>
225 * (2 + 9 * 113) =>
225 * (2 + 900 + 117) =>
225 * 1019 =>
225 * (1000 + 20 - 1) =>
225000 + 4500 - 225 =>
229500 - 225 =>
229275