Question

If `a_1,a_2,...a_n` are in H.P then the expression `a_1a_2+a_2a_3+....+a_ (n-1)a_n` is equal to
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 1

Correct Answer - D
Since `a_(1),a_(2),a_(3), . . . ,a_(n)` are in H.P.
`:." "(1)/(a_(1)),(1)/(a_(1)),(1)/(a_(3)), . . .,(1)/(a_(n))` are in A.P.
Let d be the common difference of the A.P. Then,
`(1)/(a_(2))-(1)/(a_(1))=d,(1)/(a_(3))-(1)/(a_(2))=d, . . . ,(1)/(a_(n))-(1)/(a_(n-1))=d`
`rArr" "a_(1)-a_(2)=d(a_(1)a_(2)),a_(2)-a_(3)=d(a_(2)a_(3))`,
`a_(3)-a_(4)=d(a_(3)a_(4)), . . . a_(n-1)-a_(n)=d(a_(n-1)a_(n))`
`rArr" "(a_(1)-a_(2))+(a_(2)-a_(3))+ . . . +(a_(n-1)-a_(n))`
`=d(a_(1)a_(2)+a_(2)a_(3)+ . . . +a_(n-1)a_(n))`
`rArr" "a_(1)-a_(n)=d(a_(1)a_(2)+a_(2)a_(3)+ . . . +a_(n-1)a_(n))` . . . (i)
Now,
`(1)/(a_(1)),(1)/(a_(2)),(1)/(a_(3)), . . . ,(1)/(a_(n))` are in A.P. with common difference d.
`rArr" "(1)/(a_(n))=(1)/(a_(1))+(n-1)d`
`rArr" "(1)/(a_(n))-(1)/(a_(1))(n-1)d`
`rArr" "(a_(1)-a_(n))/(a_(1)a_(n))=(n-1)drArra_(1)-a_(n)=(n-1)d(a_(1)a_(n))` . . . (ii)
From (i) and (ii), we get
`(n-1)d(a_(1)a_(n))=d(a_(1)a_(2)+a_(2)a_(3)+ . . . .+a_(n-1)a_(n))`
`rArr" "(n-1)a_(1)a_(n)=a_(1)a_(2)+a_(2)a_(3)+ . . .+a_(n-1)a_(n)`