How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have at least 3 elements?

Question

Consider the set S={ 1,2,3.....1000}. How many arithmetic progressions can be formed from the elements of S that start with 1 and end with 1000 and have atleast elements?

$An=a+(n-1)d$

$1000=1+(n-1)d$

$999=(n-1)d$

How to find the number of arithmetic progression that have atleast 3 elements?

Answer 1

$(n-1)d=999$

$d$ must be a divisor of $999$ so it may be one of the set

$D=\{1, 3, 9, 27, 37, 111, 333, 999\}$

with the first $7$ you have at least $3$ elements

for instance if $d=333$ the sequence is $1,\;334,\;667,\;1000$

with $d=999$ it doesn't work.

So the answer is $7$

Hope it helps