Question

Consider the set $S = \{1, 2, 3, \dots, 1000\}$. 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?

• A. 3
• B. 4
• C. 6
• D. 7
• E. 8

Hint:

Number of APs is linked to divisor count of the gap between first and last term.

Answer 1

The Correct Option is C

To find the number of arithmetic progressions (APs) from the set \( S = \{1, 2, 3, \dots, 1000\} \) that start with 1, end with 1000, and have at least 3 elements, follow these steps:

1. An arithmetic progression can be defined by its first term \( a \), common difference \( d \), and number of terms \( n \). The general term \( T_n \) of an AP is given by:
   \( T_n = a + (n-1) \cdot d \)
2. Since the first term \( a = 1 \) and the last term \( T_n = 1000 \), we have:
   \( 1000 = 1 + (n-1) \cdot d \)
3. Rearrange to find:
   \( (n-1) \cdot d = 999 \)
4. To ensure there are at least 3 terms, \( n \geq 3 \). For each possible \( n \), we find \( d \) such that \( 999 \) is divisible by \( n-1 \). Then:

   • \( n=3 \): \( n-1=2 \), \( d=\frac{999}{2}=499.5 \) (Not an integer)
   • \( n=4 \): \( n-1=3 \), \( d=\frac{999}{3}=333 \) (Valid AP)
   • \( n=5 \): \( n-1=4 \), \( d=\frac{999}{4}=249.75 \) (Not an integer)
   • \( n=6 \): \( n-1=5 \), \( d=\frac{999}{5}=199.8 \) (Not an integer)
   • \( n=7 \): \( n-1=6 \), \( d=\frac{999}{6}=166.5 \) (Not an integer)
   • \( n=8 \): \( n-1=7 \), \( d=\frac{999}{7}=142.714 \) (Not an integer)
   • \( n=9 \): \( n-1=8 \), \( d=\frac{999}{8}=124.875 \) (Not an integer)
   • \( n=10 \): \( n-1=9 \), \( d=\frac{999}{9}=111 \) (Valid AP)
   • \( n=12 \): \( n-1=11 \), \( d=\frac{999}{11}=90.818 \) (Not an integer)
   • \( n=14 \): \( n-1=13 \), \( d=\frac{999}{13}=76.846 \) (Not an integer)
   • \( n=19 \): \( n-1=18 \), \( d=\frac{999}{18}=55.5 \) (Not an integer)
   • \( n=28 \): \( n-1=27 \), \( d=\frac{999}{27}=37 \) (Valid AP)
   • \( n=37 \): \( n-1=36 \), \( d=\frac{999}{36}=27.75 \) (Not an integer)
   • \( n=133 \): \( n-1=132 \), \( d=\frac{999}{132}=7.568 \) (Not an integer)
   • \( n=334 \): \( n-1=333 \), \( d=\frac{999}{333}=3 \) (Valid AP)

Thus, the valid values of \( n \) and their corresponding \( d \) are:

n	d
4	333
10	111
28	37
334	3

Therefore, there are 4 possible arithmetic progressions.