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# How such A.P subsets are there???

Home Forums Math Olympiad, I.S.I., C.M.I. Entrance How such A.P subsets are there???

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• #27966
swastik pramanik
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Let $S=\{1, 2,3,\cdots, n\}$.  Find the number of subsets $A$ of $S$ satisfying the following conditions:

• $A=\{a, a+d, \cdots, a+kd\}$ for some positive integers $a, d, k$, and
• $A\cup \{x\}$ is no longer an A.P with common difference $d$ for each $x\in S\backslash A$.

(Note that $|A|\geq 2$  and any sequence of two terms is considered as an A.P).

#28159
Camellia Ray
Participant

In the given question it has been asked to find the no of subsets which satisfy the condition that
A={a, a+d,a+2d….a+kd} &

A U{x}  is no longer an A.P where x ∈ (S-A)

From above 2 condns it can b concluded that at a time with c.d d we have to form the largest sequence having c.d d and starting element say a

For ex if X={1,2,3,,,,12} then for

c.d 1:  only 1 subset possible; c.d 2 ->2 :subset possible namely { 1,3,5,7,,,11} & {2,4,6,8,,,,12}

Now c.d 1 and c.d 11 only 1 subset; c.d 2 &10 ->2 subsets

GENERALIZATION

For N Odd

1. c.d 1 & n-1 -> 1 subset;  2. c.d 2 & n-2 -> 2 subsets

similarly for c.d (n-1)/2) & cd (n+1)/(2) it will be (n-1)/2 subsets

SO TOTAL IS 2× (1+2+3+…..+(n-1)/2)

FOR EVEN

SAME as previous just one more term n/2 added

so here it is 2×(1+2+….(n−1)/2)+n/2

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