Given that,
$S = 1,2,3,\dots , 1000$
We know that, Last term of an $\text{AP} : l = a+(n-1)d,$ where $\; a=$ first term, $d=$Common difference, $n=$number of terms. $(n\ge3)$
$\Rightarrow 1000 = 1+(n-1)d$
$\Rightarrow 999 = (n-1)d$
$\Rightarrow (n-1)d= 3^{3} \times 37$
Possible values of $n-1:$
- $3$
- $3^{2}$
- $3^{3}$
- $37$
- $3 \times 37$
- $3^{2} \times 37$
- $3^{3} \times 37$
Total $7$ arithmetic progression can be formed and have at least $3$ elements.
Correct Answer $:\text{D}$