## Saturday, 26 November 2011

### Progressions -11 (CAT-2006)

Consider the series 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?
(1) 3                (2) 4                (3) 6                (4) 7                (5) 8
Solution follows here:
Solution:
“Arithmetic progressions shall start with 1 and end with 1000 and have at least 3 elements
First element = a = 1
Number of elements n≥3
Last element = tn= 1000
Let the common difference = d
In AP, tn = a+(n-1)d => 1000 = 1+(n-1)d => (n-1)*d = 999 = 33 * 371
=> 3, 9, 27, 37 etc.. are different factors of the product “(n-1)*d”
=> “(n-1)*d” can be expressed as product of two numbers in the following ways:
1*999;            999*1;            3*111;            111*3;            9*111;            111*9
27*37;            37*27;
By equating (n-1)*d to each one, we get different n and d values and hence different arithmetic progressions. But n-1 cannot be made equal to 1 as n should be greater than or equal to 3. Hence first possibility 1*999 is ruled out.
Excepting that one, all other 7 products are possible and hence 7 arithmetic progressions can be generated.