#URGENT ARITHMETHIC SEQUENCE PROBLEM
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what is arithmetic sequences
what
like 2, 4, 6, 8?
does 1, 4, 9, 16 count?
or 17,20,23
no
the difference between a term and the previous term should be the same
yea continue cooking
and you are asking me ohww many arithmetic sequences of positive integers a, b, c that satisfy a < b < c <= 1000
indeed
also yea
a < a+d < a+2d <= 1000
right noe the firdt methkd that comes to mind:
a = 1:
2d <= 999
d <= 499.5
d takes all positive integers less than, including 499.
Similarly 2:
d <= 499
d takes all positive integrrs less than and including 499
a = 3:
d <= 997/2 = 498.5
You start to realise there's a pattern. Do you see it?
so then is answer 249500?
yeah it decreases 0.5 every time you increase a by 1
I mean I guess it should be 2 times the sum i didnt really calculate anything yhis is just what came to mind
2(499 + 498 + ... + 1) or smt
Yeah thats that
basically it
yk the formula yeah
wait can u explain ur method again
i got the answer by taking d=1,2,3,.......
and for 1 i got 998, 2 i got 996 so i j continued that to 499
Take a=1
I want a + 2d <= 1000
So 1 + 2d <= 1000
2d <= 999
d <= 499.5
idk how to prove that this pattern is always the case
Now take a=2 and so on
is it enough to j say
i see pattern
hence its the answer?
No you are incrementing a by 1, thus decreasing 1000 by 1, thus your d is decreasing by intervals of 0.5
But since you want integer
You take floor of the number
Floor(4.3) = 4
Floor(4) = 4
the legendary floor
(that existed from scratch)
(Incase you dont know what floor means)
the pattern goes like 499.5
floor of it would be 499
then 499
498.5
and blah blah blah
so that's why you need to multiply by 2
ohhh
you can also take a b = half of 1000 = 500
so a = 499, 488 and so on
and c goes from 501,502 and so on
then for (a,c) it can exist as (c,a)
then for b=499
so on till b=2
where a>0
isnt this just stars and bars anyway?
a + 2d <= 1000
2d <= 1000 - a
d <= 500 - a/2
a is increasing by 1, thus a/2 is increasing by 1/2 = 0.5.