Arithmetic progression Puns

https://reddit.com/link/n7tidh/video/jc9i0ynogxx61/player

Those who don't know history are destined to repeat it. - George Santayana

Whereas China be like : Naah, they never learn, (Lets do it again) Years: 2017, 2019 ,2021 ( Nice Arithmetic Progression though)

Hey guys, hope to find out here any help. Here's the given information regarding the question:

I have an arithmetic series with 109 elements.

Sum(109 elements) = 0.

I need to calculate the n which an=0.

Tried multiple things, couldn't find an answer. Thanks.

For example: (2x+2)+(3x+5). 2 and 5 are the first terms - a, 2 & 3 are the common difference, of these 2 different Arithmetic Progressions, respectively.

Hi! I was browsing the curriculums and lists of courses of some HS , and was wandering why do we always study functions before studying arithmetic Progressions? even though it would be easier to do the opposite as Guauss "found" in his childhood doing calculus Could we teach students Arithmetic Progressions before functions by going through calculus and following the logic Guauss used? if yes and example would be helpful and thank you all

For example say: 5n+2, 5=d, n = x-1 (for any x), d = 2. So, we can know that 3,5,8, etc. are definitely not terms in this progression, by hit and trial. But, is there any direct way by which can represent the terms which are NOT any term in the Arithmetic Progression?? The particular progression I'm looking for is:

3n+1 + (x-1)(2n+1). I want to know the numbers which are not a part of this progression. Thank you

My school skipped over the topic because we ran late on the syllabus last year and I need to make sure I don't miss anything because my math subject test is in 2 weeks. Can anyone help?

Hi, I'm doing some exercises where I need to find which line a certain number is, like:

First Line - 1

Second Line - 2 3 4

Third Line - 5 6 7 8 9

Fourth Line -> 10 11 12 13 14 15 16

Fifth Line -> 17 18 19 20 21 22 23 24 25

Nth line - (...) 1969 (...)

and I need to find which line the number 1969 is, so I thought about arithmetic progression but I don't know how to do it

Not sure if it's already well-known but I have been exploiting this "bug" for a while.

When you keep prisoners in your party, in time they will gradually become recruitable. There will be an icon that indicates how many prisoners of this unit you can recruit. For example, if you have 10 imperial trained infantryman and the icon says 4, technically it means you can left click icon 4 times and recruit 4 units with 6 prisoners left. However, if you do it by holding SHIFT then left click it 4 times, you can actually recruit 4 + 3 + 2 + 1 = 10 units. You can recruit all of them! Each SHIFT click gives you "x" unit of the prisoners (where x is the face value of the icon) while it only reduces its number by 1 even with SHIFT, so you can SHIFT click it 4 times.

SHIFT click 2 times can recruit 3 (2 + 1), while normal click 2 times only recruit 2 (1 + 1) with 1 prison left

On the other hand, you can also keep accumulating the recruitable number and only use one single SHIFT click and then wait for the number to come back (the recruitable number is of course capped at the amount of the prisoners with the same type in your party).

I usually keep like 4~5 prisoners of the units I want in my party and wait for the icon to accumulate. After that I go to my settlement and pull more prisoners with the same type then use this bug to recruit many of them. This is especially useful when you want to recruit a plenty of high-tier prisoners.

Remember this trick only works when you are holding SHIFT and left click. It works normally (+1, +1, ...) if you just left click without SHIFT.

In conclusion, don't recruit prisoners as soon as they become available. Keep these recruitable in your party and accumulate the recruitable number. Then use SHIFT click trick to recruit a plenty of them!

The first term of an arithmetic progression | is the same as the 4th term of an arithmetic progression ||which has a common difference 2. Find (a) The first term of the arithmetic progression || (b) The common difference of the arithmetic progression | if the sum of the first four term of this progression is 22

I feel like this is a simple question but I just can't do it... I've spent more time on this than I care to admit HALP

First i have to say that i'm totally not a specialist in number theory so maybe this post talks about something that isn't special at all.

I was interested in a recent post which aimed at finding a number that is prime when it's value in base 10 was interpreted in all other bases under 10. I started writing the number 101 and calculated the value in every bases from 2 to 10. I was very surprised to see that from base 2 to 3 you just had to add 5, from base 3 to 4 add 7, from 4 to 5 add 9, and the pattern holds until base 10. For short, the number 1101 in every base was the sequence 2n+5. I tried with 1011 but there wasn't any arithmetic progression. But the numbers you add from a base to another were folowing the arithmetic progression 6n+18 So it means with more digits i had to go a "level" deeper to find an arithmetic progression.

To sum up, i found that if you take the difference of consecutive numbers in a sequence it is operation D (for difference) For a given number about n digits long, i found examples where if you have the sequence of all bases interpretations of that number as sequence U Then if you apply n-2 times D to this sequence you get an arithmetic progression.

This works for 1, 11, 101, 1011, 10111 but i didn't try more. Since my computer is broken i can't do any programming for that right now.

Tell me what you think of this and if you have spare time, enjoy yourself making a program to verify if this always work.

Thank you for reading this long post

I have the following progressions:

a1, a2, a3 and a4 make an arithmetic progression.

(a1-2), a2, (a3+10) and (a4+36) make a geometric progression.

The question is: Which are the numbers that make up both progressions?

Those who don't know history are destined to repeat it. - George Santayana

Whereas China be like : Naah, they never learn, (Lets do it again) Years: 2017, 2019 ,2021 ( Nice Arithmetic Progression though)