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if `a,a_1,a_2,a_3,.........,a_(2n),b` are in `A.P.` and `a,g_1,g_2,............g_(2n) ,b` are in `G.P.` and `h` is `H.M.` of `a,b` then `(a_1+a_(2n))/(g_1*g_(2n))+(a_2+a_(2n-1))/(g_2*g_(2n-1))+............+(a_n+a_(n+1))/(g_n*g_(n+1))` is equalA. `(2n)/(h)`B. 2nhC. nhD. `(n)/(h)`

Answer» Correct Answer - A
We have,
`a_(1)+2_(2n)=a_(2)+a_(2n-1)+ . . . =a_(n)+a_(n+1)=a+b`
and,
`g_(1)g_(2n)=g_(2)g_(2n-1)= . . .=g_(n)g_(n+1)=ab`
`:." "(a_(1)+a_(2n))/(g_(1)g_(2n))+(a_(2)+a_(2n-1))/(g_(2)g_(2n-1))+ . . . .+(a_(n)+a_(n+1))/(g_(n)g_(n+1))`
`=n((a_(1)+a_(2n))/(g_(1)g_(2n)))=((a+b)/(ab))=(2n)/(h)" "{:[becauseh=(2ab)/(a+b)]:}`


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