IIT JEE , AIEEE Forum - Mathematics , Algebra - A Very Good Question with a salute waiting

kghedriu

If n=p, the ans is trivial......

suppose n=/= p...

we know that Sk=Ak^2+Bk where A=d/2 and B=a-d/2

from given conditions...

2(Am^2+Bm)=A (m+n)^2+ B(m+n)

=>A(m^2-2mn-n^2)+B(m-n)=0.................................(1)

also, A(m^2-2mp-p^2)+B(m-p)=0.............................(2)

so, A(-2mp+2mn+n^2-p^2) + B(n-p)=0

=> A(2m+n+p)+B=0.....................................................(3) since n=/= p

from (1) and (3),

2=(m+p)(1/m-1/n)

from (2) and (3),

2=(m+n)(1/m-1/p)

so, (m+p)(1/m-1/n)=(m+n)(1/m-1/p)

hence the answer...