Show that 1a1a2-1a2a3- -1an-1an-n-1a1an-

1a_1 1a_2=a_2 a_1a_1a_2=da_1a_2 ∴ d∑1a_1a_2=∑(1a_1 1a_2) The terms will cancel diagonally =1a_1 1a_n=a_n a_1a_1a_n d∑1a_1a_2=(n 1)da_1a_n ...

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Angle between Two Lines

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Question

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

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

tan [{tan}^{- 1} (\frac{d}{1 + a_{1} a_{2}}) + {tan}^{- 1} (\frac{d}{1 + a_{2} a_{3}}) + . . . + {tan}^{- 1} (\frac{d}{1 + a_{n - 1} a_{n}})] = \frac{(n - p) d}{q + a_{1} a_{n}}

p + q

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