- 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}\)
.
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