If `a_(1),a_(2),".....",` are in HP and `f_(k)=sum_(r=1)^(n)a_(r)-a_(k)`, then `2^(alpha_(1)),2^(alpha_(2)),2^(alpha_(3))2^(alpha_(4)),"....."` are in `{" where " alpha_(1)=(a_(1))/(f_(1)),alpha_(2)=(a_(2))/(f_(2)),alpha_(3)=(a_(3))/(f_(3)),"....."}`.

Options

A. AP
B. GP
C. HP
D. None of these

Answer

D

Explanation

`a_(1),a_(2),a_(3),"......"` are in HP.
`implies (1)/(a_(1)),(1)/(a_(2)),(1)/(a_(3)),"......" " are in AP. " " " "......(i)"`
`:.f_(k)=sum_(r=1)^(n)a_(r)-a_(k)`
`implies a_(k)+f_(k)=sum_(r=1)^(n)a_(r)=lambda " " [" say "]`
`impliesa_(1)+f_(1)=a_(2)+f_(2)=a_(3)+f_(3)="....."=lambda`
From Eq. (i), `(lambda)/(a_(1)),(lambda)/(a_(2)),(lambda)/(a_(3))"......"` are in also in AP.
`(a_(1)+f_(1))/(a_(1)),(a_(2)+f_(2))/(a_(2)),(a_(3)+f_(3))/(a_(3)),"......"` are also in AP.
Subtracting from each term by 1, we get
`(f_(1))/(a_(1)),(f_(2))/(a_(2)),(f_(3))/(a_(3)),"......"` are also in AP.
`(1)/(alpha_(1)),(1)/(alpha_(2)),(1)/(alpha_(3)),"....."` are in AP.
`:.(1)/(alpha_(1)),(1)/(alpha_(2)),(1)/(alpha_(3)),"....."` are in HP.
`2^(alpha_(1)),2^(alpha_(2)),2^(alpha_(3)),"....."` are not in AP/GP/HP.