If $a_{1}, a_{2}, \ldots, a_{n}$ are in H.P., then the expression $a_{1} a_{2}+a_{2} a_{3}+\ldots+a_{n-1} a_{n}$ is equal to

JEE Main · 2006 · 30 Apr 2006 · Offline · Sequences and Series

a_{1}, a_{2}, \dots, a_{n}

are in H.P., then the expression

a_{1} a_{2} + a_{2} a_{3} + \dots + a_{n - 1} a_{n}

is equal to

Answer: (D)

a_{1}, a_{2}, \dots, a_{n}

are in H.P., their reciprocals are in A.P. with some common difference

d

, so

\frac{1}{a _{r + 1}} - \frac{1}{a _{r}} = d

, that is

\frac{a _{r} - a _{r + 1}}{a _{r} a _{r + 1}} = d

and hence

a_{r} a_{r + 1} = \frac{a _{r} - a _{r + 1}}{d}

. Summing from

r = 1

n - 1

telescopes to

a_{1} a_{2} + \dots + a_{n - 1} a_{n} = \frac{a _{1} - a _{n}}{d}

. The A.P. relation

\frac{1}{a _{n}} = \frac{1}{a _{1}} + (n - 1) d

gives

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

, so

\frac{a _{1} - a _{n}}{d} = (n - 1) a_{1} a_{n}

. The sum therefore equals

(n - 1) a_{1} a_{n}

JEE Main · 2006 · 30 Apr 2006 · Offline · Single correct option