CAT Question

Yashika

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

1) 3

2) 4

3) 6

4) 7

5) 8

$hubham... Off to a nu start!!

1000 = 1 + (n-1) dso n = 1 + 999/dn shud be integer ..same wid d(bcz no element of AP is in fraction).....so d shud be factor of 999.....which r 1,3,9,27,37,111,333....not possible further i think as elements shud be atleast 3....so 7 i think...hence option 4....

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