If `a ,a_1, a_2, a_3, a_(2n),b` are in A.P. and `a ,g_1,g_2,g_3, ,g_(2n),b` . are in G.P. and `h` s the H.M. of `aa n db ,` then prove that `(a_1+a_(2n))/(g_1g_(2n))+(a_2+a_(2n-1))/(g_1g_(2n-1))++(a_n+a_(n+1))/(g_ng_(n+1))=(2n)/h`

Options

A. `(2n)/(h)`
B. 2nh
C. nh
D. `(n)/(h)`

Answer

A

Explanation

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)]:}`