2016/91) if $a_1,a_2,\cdots,a_n$ are in HP show that $a_1a_2+a_2a_3+\cdots + a_{n-1}a_n = (n-1)a_1a_n$

we have
  $\frac{1}{a_1},\frac{1}{a_2}\cdots,\frac{1}{a_n}$ are in AP hence
or $(n-1)(\frac{1}{a_2} - \frac{1}{a_1})  = \frac{1}{a_n} - \frac{1}{a_1}$
or $(n-1)(a_1-a_2)(a_n a_1) = a_2a_1(a_1-a_n) \cdots(1)$
similarly
 $(n-1)(a_2-a_3)(a_n a_1) = a_3a_2(a_1-a_n) \cdots(2)$\
...
   $(n-1)(a_{n-1}-a_n)(a_n  a_1) = a_{n-1}a_m(a_1-a_n) \cdots(n-1)$\
adding above n eqautions we get
$(n-1)(a_1-a_n)(a_n a_1) = (a_1a_2+a_2a_3+\cdots + a_{n-1}a_n) (a_1-a_n)$
cancelling $a_1-a_n$ from both sides we get the result