If a1- a2- a3-an be an AP of non-zero terms - then 1a1a2 - 1a2a3 -1an-1an -

The correct option is C n 1a_1.a_n ...

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nth Term of A.P

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Question

If $a_{1}, a_{2}, a_{3}, . . . . . . a_{n}$ be an AP of non-zero terms , then $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + . . . . . + \frac{1}{a_{n - 1} a_{n}} =$

\frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n}}}

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\frac{n - 1}{a_{1} + a_{n}}

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\frac{n - 1}{a_{1} . a_{n}}

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\frac{2 - n}{a_{1} . a_{n}}

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Similar questions

If $a_{1}, a_{2}, a_{3}, . . . ., a_{n}$ be an AP of non-zero terms. Prove that

$\frac{1}{a_{1} a_{2}}$ + $\frac{1}{a_{2} a_{3}}$ +....+ $\frac{1}{a_{n - 1} a_{n}}$ =

$\frac{n - 1}{a_{1} a_{n}}$

If $a_{1}, a_{2}, a_{3}, \dots, a_{n}$ be an AP of nonzero terms then prove that

$\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + \dots + \frac{1}{a_{n - 1} a_{n}} = \frac{(n - 1)}{a_{1} a_{n}}$

a_{1}, a_{2}, a_{3} . . . ., a_{n}

t a n [t a n^{- 1} (\frac{d}{1 + a_{1} a_{2}})] + t a m^{- 1} (\frac{d}{1 + a_{2} a_{3}})] + . . . + t a n^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] =

a_{1}, a_{2}, a_{3}, . . ., a_{n}

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + . . . \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}

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