If `a_(1),a_(2),a_(3),".....",a_(n)` are in HP, than prove that `a_(1)a_(2)+a_(2)a_(3)+a_(3)a_(4)+"....."+a_(n-1)a_(n)=(n-1)a_(1)a_(n)`

Explanation

Given, `a_(1),a_(2),a_(3),"…",a_(n)` are in HP.
` therefore (1)/a_(1),(1)/a_(2),(1)/a_(3),"....."(1)/a_(n)` are in AP.
Let D be the common difference of the AP, than
`(1)/a_(2)-(1)/a_(1)=(1)/a_(3)-(1)/a_(2)=(1)/a_(4)-(1)/a_(3)="....."=(1)/a_(n)-(1)/a_(n-1)=D`
` implies (a_(1)-a_(2))/(a_(1)a_(2))=(a_(2)-a_(3))/(a_(2)a_(3))=(a_(3)-a_(4))/(a_(3)a_(4))="....."=(a_(n-1)-a_(n))/(a_(n-1)a_(n))=D`
` implies a_(1)a_(2)=(a_(1)-a_(2))/(D),a_(2)a_(3)=(a_(2)-a_(3))/(D),a_(3)a_(4)=(a_(3)-a_(4))/(D),".....",a_(n-1)-a_(n)=(a_(n-1)a_(n))/(D)`
On adding all such expressions, we get
` a_(1)a_(2)+a_(2)a_(3)+a_(3)a_(4)+"....."+a_(n-1)-a_(n)=(a_(1)-a_(n))/(D)=(a_(1)a_(n))/(D)((1)/a_(n)-(1)/a_(1))`
` (a_(1)a_(n))/(D)[(1)/a_(1)+(n-1)D-(1)/(a_(1))]=(n-1)a_(1)a_(n)`
Hence, ` a_(1)a_(2)+a_(2)a_(3)+a_(3)a_(4)+"....."+ a_(n-1)a_(n)=(n-1)a_(1)a_(n)`