a₁, a₂, a₃ ..... , an are in H.P.

⇒ \(\frac{1}{a_1}\),\(\frac{1}{a_2}\),\(\frac{1}{a_3}\) ......., \(\frac{1}{a_n}\) are in A.P

If d is the common difference of the A.P., then

\(\frac{1}{a_2}\) - \(\frac{1}{a_1}\) = d, \(\frac{1}{a_3}\) - \(\frac{1}{a_2}\) = d,......, \(\frac{1}{a_n}\)- \(\frac{1}{a_{n-1}}\) = d

⇒ \(\frac{a_1-a_2}{a_1a_2}\) = d,  \(\frac{a_2-a_3}{a_2a_3}\) = d, ....., \(\frac{a_{n-1}-a_n}{a_{n-1}\,a_n}\) = d

⇒ a₁ – a₂ = a₁a₂d, a₂ – a₃ = a₂a₃d, ..... , a_{n – 1} – a_n = a_{n – 1} a_nd 

⇒ (a₁ – a₂ + a₂ – a₃ + ..... + a_{n – 1} – a_n) = a₁a₂d + a₂a₃d + ..... + a_{n – 1} a_nd

⇒ (a₁ – a_n) = (a₁a₂ + a₂a₃ + ..... + a_{n – 1} a_n)d                          ....(i)

Also, \(\frac{1}{a_1}\),\(\frac{1}{a_2}\),.......\(\frac{1}{a_n}\) is an A.P with common difference d

⇒ \(\frac{1}{a_n}\) = \(\frac{1}{a_1}\) + (n - 1)d ⇒ (n - 1)d = \(\frac{1}{a_n}\) - \(\frac{1}{a_1}\) ⇒ (n - 1)d = \(\frac{a_1-a_n}{a_1a_n}\)

⇒ a₁ – a_n = (n – 1)d a₁a_n                   ...(ii)

From (i) and (ii) 

(n – 1)d a₁a_n = (a₁a₂ + a₂a₃ + .... + a_{n – 1}a_n) d 

⇒ (a₁a₂ + a₂a₃ + .... + a_{n – 1}a_n) = (n – 1) a₁a_n.