Question

If `a_1,a_2,a_3,...,a_(n+1)` are in A.P. , then `1/(a_1a_2)+1/(a_2a_3)....+1/(a_na_(n+1))` is
A. `(n-1)/(a_(1)a_(n+1))`
B. `(1)/(a_(1)a_(n+1))`
C. `(n+1)/(a_(1)a_(n+1))`
D. `(n)/(a_(1)a_(n+1))`

Answer 1

Correct Answer - D
Let d be the common difference of the given A.P.
Then,
`(1)/(a_(1)a_(2))+(1)/(a_(2)a_(2))+ . . . +(1)/(a_(n)a_(n+1))`
`=(1)/(d){(a_(2)-a_(1))/(a_(1)a_(2))+(a_(3)-a_(2))/(a_(2)a_(3))+ . . .(a_(n+1)-a_(n))/(a_(n)a_(n+1))}`
`=(1)/(d){(1)/(a_(1))-(1)/(a_(2))+(1)/(a_(2))-(1)/(a_(3))+ . . +(1)/(a_(n))-(1)/(a_(n-1))}`
`=(1)/(d)((1)/(a_(1))-(1)/(a_(n+1)))=(a_(n+1)-a_(1))/("d "a_(1)a_(n+1))=(nd)/("d "a_(1)a_(n+1))=(nd)/("d "a_(1)a_(n+1))=(n)/(a_(1)a_(n+1))`