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

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

⇒ $\frac{a_1-a_2}{a_1{a}_2}=\frac{a_2-a_3}{a_2{a}_3}=\frac{a_3-a_4}{a_3{a}_4}=....=\frac{a_{n-1}-a_n}{a_{n-1}{a}_n}=d$

⇒ $a_1{a}_2=\frac{a_1-a_2}{d},a_2{a}_3=\frac{a_2-a_3}{d},....,a_{n-1}{a}_n=\frac{a_{n-1}-a_n}{d}$

adding all the above we get.

${\mathrm{a}}_1{\mathrm{a}}_2+{\mathrm{a}}_2{\mathrm{a}}_3+....{\mathrm{a}}_{\mathrm{n}-1}{\mathrm{a}}_{\mathrm{n}}=\frac{{\mathrm{a}}_1-{\mathrm{a}}_{\mathrm{n}}}{\mathrm{d}}$

$=\frac{a_1{a}_n}{d}(\frac{1}{a_n}-\frac{1}{a_1})$

$=\frac{a_1{a}_n}{d}(\frac{1}{a_1}+(n-1)d-\frac{1}{a_1})$

$=\frac{{\mathrm{a}}_1{\mathrm{a}}_{\mathrm{n}}}{\mathrm{d}}((\mathrm{n}-1)\mathrm{d})$

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