If `a_1, a_2, a_3,.....a_n` are in H.P. and `a_1 a_2+a_2 a_3+a_3 a_4+.......a_(n-1) a_n=ka_1 a_n`, then k is equal to

Options

A. `n(a_(1)-a_(n))`
B. `(n-1)(a_(1)-a_(n))`
C. `na_(1)a_(n)`
D. `(n-1)a_(1)a_(n)`

Answer

D

Explanation

`(1)/(a_(2))-(1)/(a_(1))=(1)/(a_(3))-(1)/(a_(2))="....."=(1)/(a_(n))-(1)/(a_(n-1))=d" "[" say "]`
Then `a_(1)a_(2)=(a_(1)-a_(2))/(d),a_(2)a_(3)=(a_(2)-a_(3))/(d)".....",a_(n-1)a_(n)=(a_(n-1)-a_(n))/(d)`
`:.a_(1)a_(2)+a_(2)a_(3)+"....."+a_(n-1)a_(n)=(a_(1)-a_(n))/(d)`
Also, `(1)/(a_(n))=(1)/(a_(1))+(n-1)d implies (a_(1)-a_(n))/(d)=(n-1)a_(1)a_(n)`
`:.a_(1)a_(2)+a_(2)a_(3)+"......"+a_(n-1)a_(n)=(n-1)a_(1)a_(n)`.