Home › Forums › Math Olympiad, I.S.I., C.M.I. Entrance › Solution for how such a.p subsets r there

- This topic has 0 replies, 1 voice, and was last updated 10 months, 1 week ago by Camellia Ray.

- AuthorPosts
- May 27, 2019 at 3:00 am #28122Camellia RayParticipant
- 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 (\in) 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 nd 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 nd c.d 11 only 1 subset;

c.d 2&10 ->2 subsets

GENERALIZATION

For n odd

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

c.d 2& n-2 -> 2

similarly for c.d\ (\frac{n-1}{2})

& cd (\frac{n+1}{2}) it will b

e\ (\frac{n-1}{2})

SO TOTAL IS 2× (1+2+3+…..

(\frac{n-1}{2}))

FOR EVEN

SAME as previous just one more

term (\frac{n}{2}) added

so here it is 2×(1+2+….

- \(\frac{n-1}{2}\))+\ (\frac{n}{2}\)

- This topic was modified 10 months, 1 week ago by Camellia Ray.
- This topic was modified 10 months, 1 week ago by Camellia Ray.

###### Attachments:

You must be logged in to view attached files. - AuthorPosts

- You must be logged in to reply to this topic.