Solution for how such a.p subsets r there

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    Camellia Ray
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    1. 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 2 months, 4 weeks ago by  Camellia Ray.
    • This topic was modified 2 months, 4 weeks ago by  Camellia Ray.
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