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CAT-2006-Previous Years Paper

Question
14 out of 25
 

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 at least 3 elements?



A 3

B 4

C 6

D 7

E 8

Ans. D

Let number of terms in progression be n, then using tn = a + (n – 1) d

1000 = 1 + (n – 1)d, or, (n – 1)d = 999 = 33 × 37.

Now, making a table of values of n and d gives us the following values of (n-1):

(n –1) = 3 or 9 or 27 or 37 or 111 or 333 or 999

Hence, 7 progressions are possible.

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