if the first and (2n_1)th term of an AP,GP andHP are equal and their nth term are a,b,and c respectively,then

dear sudhir singh

let first term is x

then (2n-1)th term

 x+(2n-2)d₁  = xr^2n-2 = x/{1+(2n-2)d₂} =k

so  (n-1)d1=(k-x)/2

and  r^n-1=√k/x

and   x (n-1)d2 =(x-k)/2k

so for nth term

given   a = x+(n-1)d1

              =(k+x)/2

         b = xr^n-1 = √kx

        c = x/({1+(n-)d₂} = 2kx/(x+k)

clearly   a.c =b²

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