If $a_{1},\ a_{2},\ a_{3},\ \ldots.,\ a_{n}$ are in AP with common ratio $d$, then $ an\left\lbrack an^{- 1}\ \frac{d}{1 + a\_ 1a_{2}} + an^{- 1}\ \frac{d}{1 + a_{2}a_{3}} + ... + an^{- 1}\ \frac{4}{1 + a_{n - 1}a_{n}} ight brack$ is equal to

Question

If $a_{1},\ a_{2},\ a_{3},\ \ldots.,\ a_{n}$ are in AP with common ratio $d$, then $	an\left\lbrack 	an^{- 1}\ \frac{d}{1 + a\_ 1a_{2}} + 	an^{- 1}\ \frac{d}{1 + a_{2}a_{3}} + ... + 	an^{- 1}\ \frac{4}{1 + a_{n - 1}a_{n}} ightbrack$ is equal to

Dr. Abubacker · Accepted Answer

(b)
Given expression
$= 	an\left\lbrack 	an^{- 1}{\frac{a_{2} - a_{1}}{1 + a_{1}a_{2}} + 	an^{- 1}{\frac{a_{3} - a_{2}}{1 + a_{2}a_{3}} + ... + 	an^{- 1}\frac{a_{n} - a_{n - 1}}{1 + a_{n}a_{n - 1}}}} ightbrack$
$= 	an{\lbrack	an^{- 1}{a_{2} - 	an^{- 1}{a_{1} + 	an^{- 1}{a_{3} - 	an^{- 1}{a_{2} + ... + 	an^{- 1}{a_{n} - 	an^{- 1}a_{n - 1}}}}}}}$]
$= tan\lbrack	an^{- 1}{a_{n} - 	an^{- 1}{a_{1}brack = \frac{a_{n} - a_{1}}{1 + a_{1}a_{n}}}}$
$= \frac{(n - 1)d}{1 + a_{1}a_{n}}$

Q. If $a_{1},\ a_{2},\ a_{3},\ \ldots.,\ a_{n}$ are in AP with common ratio $d$, then $\tan\left\lbrack \tan^{- 1}\ \frac{d}{1 + a\_ 1a_{2}} + \tan^{- 1}\ \frac{d}{1 + a_{2}a_{3}} + ... + \tan^{- 1}\ \frac{4}{1 + a_{n - 1}a_{n}} \right\rbrack$ is equal to

Explanation

Reviewed by Dr. Abubacker

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