1.

Let `a_1, a_2, a_3, ...a_(n)` be an AP. then: `1 / (a_1 a_n) + 1 / (a_2 a_(n-1)) + 1 /(a_3a_(n-2))+......+ 1 /(a_(n) a_1) = `A. 2B. `a_(1)_+a_(n)`C. `2(a_(1)+a_(n))`D. `(2)/(a_(1)+a_(n))`

Answer» Correct Answer - D
We know that the sum of the terms equidistant from the beginning and the end an A.P. is always same i.e.
`a_(1)+a_(n)=a_(2)+a_(n-1)=a_(3)+a_(n-2)= . . .=a_(k)+a_(n-(k+1))= . . .`
`:." "(a_(1)+a_(n)){(1)/(a_(1)a_(n))+(1)/(a_(2)a_(n-1))+(1)/(a_(3)a_(n-1))+ . . . +(1)/(a_(n)a_(1))}`
`=(a_(1)+a_(n))/(a_(1)a_(n))+(a_(1)+a_(n))/(a_(2)a_(n-1))+(a_(1)+a_(n))/(a_(3)a_(n-1))+ . . .+(a_(1)+a_(n))/(a_(n)a_(1))`
`=(a_(1)+a_(n))/(a_(1)a_(n))+(a_(2)+a_(n-1))/(a_(2)a_(n-1))+(a_(3)+a_(n-2))/(a_(3)a_(n-2))+ . . . (a_(1)+a_(n))/(a_(1)a_(n))`
`=((1)/(a_(n))+(1)/(a_(1)))+((1)/(a_(n-1))+(1)/(a_(2)))+((1)/(a_(n-2))+(1)/(a_(3)))+. . . +((1)/(a_(n-1))+(1)/(a_(2)))+((1)/a_(n)+(1)/(a_(1)))`
`=2{(1)/(a_(1))+(1)/(a_(2))+(1)/(a_(3))+ . . .(1)/(a_(n))}`
`:." "(1)/(a_(1)a_(n))+(1)/(a_(2)a_(n-1))+(1)/(a_(3)a_(n-2))+ . . . +(1)/(a_(n)a_(1))`
`=(2)/(a_(1)+a_(n)){(1)/(a_(1))+(1)/(a_(2))+ . . .+(1)/(a_(n))}`
`:." "lamda=(2)/(a_(1)+a_(n))`


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