If a1- a2 -a3 -an are in AP - then 1a1a2-1a2a3-1a3a4-1an-1an equals-

The correct option is D n 1a_1a_n If a_1, a_2, a_3,...........................an 1, a_n are in A.Pthen common difference d=a_n an 1=........=a_3 ...

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Sum of First n Natural Numbers

If a1, a2 ,...

Question

If $a_{1}, a_{2}, a_{3}, . . . .,$ an are in AP , then $\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} + \frac{1}{a_{3} a_{4}} + . . . . . + \frac{1}{a_{n - 1} a_{n}}$ equals.

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

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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{n - 1}{a_{1} a_{n}}

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

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

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

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

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

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

a_{1} a_{2} + a_{2} a_{3} + a_{3} a_{4} + . . . . + a_{n - 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}})] =

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