If a₁, a₂, a₃, ........ are in H.P., then the expression a₁a₂ + a₂a₃ +...... + a_{n - 1} a_n is equal to:

Calculations:

Given that a1, a2, a3,....an are in H.

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

Let d be the common difference of the A.P., then

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

⇒ \(\frac{a_1-a_2}{a_1a_2}=\frac{a_2-a_3}{a_2a_3}=\frac{a_3-a_4}{a_3a_4}=....=\frac{a_{n-1}-a_n}{a_{n-1}a_n}=d\)

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

adding all the above we get.

\(\rm a_1a_2+a_2a_3+....a_{n-1}a_n=\frac{a_1-a_n}{d}\)

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

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

\(\rm =\frac{a_1a_n}{d}((n-1)d)\)

∴ a1a2 + a2a3 + .....+ an-1an = (n - 1)a1an