If `s_1,s_2,s_3` denote the sum of n terms of 3 arithmetic series whose first terms are unity and their common difference are in H.P., Prove that `n=(2s_3s_1-s_1s_2-s_2s_3)/(s_1-2s_2+s_3)`

Explanation

Let `d_(1),d_(2)" and "d_(3)` be the common differences of the 3 arithmetic progressions.
`:. S_(i)=(n)/(2)[2xxa+(n-1)d_(i)],AA_(i)=1,2,3`
`implies S_(i)=(n)/(2)[2+(n-1)d_(i)]`
`implies S_(i)=n+(n(n-1))/(2)d_(i) implies d_(i)=(2(S_(i)-n))/(n(n-1))`
Given that `d_(1),d_(2),d_(3)` are in HP.
`:. (1)/(d_(1)),(1)/(d_(2)),(1)/(d_(3))` are in AP.
`:. (2)/(d_(2))=(1)/(d_(1))+(1)/(d_(3))`
`implies (2)/((2(S_(2)-n))/(n(n-1)))=(1)/((2(S_(1)-n))/(n(n-1)))+(1)/((2(S_(3)-n))/(n(n-1)))`
`implies (2)/(S_(2)-n)=(1)/(S_(1)-n)+(1)/(S_(3)-n)`
`implies (2)/(S_(2)-n)=(S_(3)+S_(1)-2n)/((S_(1)-n)(S_(3)-n))`
`implies 2[S_(1)S_(3)-(S_(1)+S_(3))n+n^(2)]=(S_(2)-n)(S_(1)+S_(3)-2n)`
`implies 2S_(1)S_(3)-2(S_(1)+S_(3))n+2n^(2)`
`implies S_(1)S_(2)+S_(2)S_(3)-2nS_(2)-n(S_(1)+S_(3))+2n^(2)`
`implies 2S_(1)S_(3)-S_(2)S_(3)-S_(1)S_(2)=n(S_(1)+S_(3)-2S_(2))`
`implies n=((2S_(1)S_(3)-S_(2)S_(3)-S_(1)S_(2)))/((S_(1)-2S_(2)+S_(3)))`.