Progression - no.of AP's possible

Question

Let S = {1,2,3,….,541}. Find out how many AP’s can be formed by taking the elements of S starting from 1 and ending at 541. Which have atleast 3 elements.

A – 12;      B – 13;      C – 23;      D – 24;      E – 27      

Answer 1

basic counting

I took difference 541-1=540 this is done to find how many common difference is possible

now take factors of 540=1*2*2*3*3*3*5

so now we take all possible combination which we can take as common difference and which ensure atleast 3 terms.

{1,2,3,4,5,6,9,10,12......,180,270}= total 24 terms hence answer is D

Comments

C option : 23 

Assume AP as 1,1+d,1+2d,.....,541
So 1+(n-1)d = 541
(n-1)d = 540.
So we have to find the number of factors of 540 such that n >= 3 or n-1 >= 2 because the minimum number of terms in AP is given to be 3. 
The number of factors of 540 is 24 but we have to exclude the case of n=2 and d=540 as it will form an AP of 2 terms. Hence we will get 23 A.P

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Why you excluded case where d=1 ?

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I excluded the case of n=2 and d=540 which means only 2 terms but in the question it is mentioned that atleast 3 terms.

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1 is also an factor and you seems to miss that