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 ∈ (SA)
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
 c.d 1 & n1 > 1 subset; 2. c.d 2 & n2 > 2 subsets
similarly for c.d (n1)/2) & cd (n+1)/(2) it will be (n1)/2 subsets
SO TOTAL IS 2× (1+2+3+…..+(n1)/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

This reply was modified 2 months, 4 weeks ago by Camellia Ray.

This reply was modified 2 months, 4 weeks ago by Camellia Ray.

This reply was modified 2 months, 4 weeks ago by Camellia Ray.

This reply was modified 2 months, 4 weeks ago by Camellia Ray.

This reply was modified 2 months, 4 weeks ago by Camellia Ray.