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Consider the set $\text{S} = \left \{ 1, 2, 3, \dots,1000 \right \}$. How many arithmetic progressions can be formed from the elements of $\text{S}$ that start with $1$ and end with $1000$ and have at least $3$ elements?

  1. $3$
  2. $4$
  3. $6$
  4. $7$
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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 $:\text{D}$

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