If a1-a2- a3- - - an are in AP with common difference 5 and if aiaj - 1 for i- j - 1- 2- - - N- then tan-15- -1-a1a2- - tan-15-1 - a2a3- - - - - tan-15-1 - an-1an- -

Explanation for the correct option:Finding the value of 𝑡𝑎𝑛^ 15/[1+𝐚_1𝐚_2]+𝑡𝑎𝑛^ 15/[1+𝐚_2𝐚_3]+….+𝑡𝑎𝑛^ 15/[1+𝐚_𝐧 1𝐚_𝐧]:Given a_1,a_2, a_3, … .are in AP with com ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Integration Using Substitution

If a1,a2, a3,...

Question

If $a_{1}, a_{2}, a_{3}, \dots .$ an are in AP with common difference $5$ and if $a_{i} a_{j} \neq - 1 f o r i, j = 1, 2, \dots . N,$ then $\tan^{- 1} \frac{5}{[1 + a_{1} a_{2}]} + \tan^{- 1} \frac{5}{[1 + a_{2} a_{3}]} + \dots . + \tan^{- 1} \frac{5}{[1 + a_{n - 1} a_{n}]} =$

$\tan^{- 1} \frac{5}{[1 + a_{n - 1} a_{n}]}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\tan^{- 1} \frac{5 a_{1}}{[1 + a_{n} a_{1}]}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\tan^{- 1} \frac{5 n - 5}{[1 + a_{n} a_{1}]}$

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

$\tan^{- 1} \frac{5 n - 5}{[1 + a_{n + 1} a_{1}]}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

$\tan^{- 1} \frac{5 n}{[1 + a_{n} a_{1}]}$

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C
$\tan^{- 1} \frac{5 n - 5}{[1 + a_{n} a_{1}]}$

Explanation for the correct option:
Finding the value of $t a n^{- 1} \frac{5}{[1 + a_{1} a_{2}]} + t a n^{- 1} \frac{5}{[1 + a_{2} a_{3}]} + \dots . + t a n^{- 1} \frac{5}{[1 + a_{n - 1} a_{n}]} :$
Given $a_{1}, a_{2}, a_{3}, \dots .$ are in AP with common difference $d = 5$
$a_{2} - a_{1} = a_{3} - a_{2} = a_{4} - a_{3} = . . . . . = a_{n} - a_{n - 1} = 5 \tan^{- 1} \frac{5}{[1 + a_{1} a_{2}]} + \tan^{- 1} \frac{5}{[1 + a_{2} a_{3}]} + \dots . + \tan^{- 1} \frac{5}{[1 + a_{n - 1} a_{n}]} = \tan^{- 1} \frac{{(a_{2} - a}_{1})}{[1 + a_{1} a_{2}]} + \tan^{- 1} \frac{(a_{3} - a_{2})}{[1 + a_{2} a_{3}]} + \dots . + \tan^{- 1} \frac{(a_{n} - a_{n - 1})}{[1 + a_{n - 1} a_{n}]} = \tan^{- 1} a_{2} - \tan^{- 1} a_{1} + \tan^{- 1} a_{3} - \tan^{- 1} a_{2} + . . . . \tan^{- 1} a_{n} - \tan^{- 1} a_{n - 1} = \tan^{- 1} a_{n} - \tan^{- 1} a_{1} = \tan^{- 1} \frac{(a_{n} - a_{1})}{[1 + a_{1} a_{n}]} = \tan^{- 1} \frac{(a_{n} - a_{n - 1})}{[1 + a_{n - 1} a_{n}]} = \tan^{- 1} \frac{5 (n - 1)}{[1 + a_{n - 1} a_{n}]} = \tan^{- 1} \frac{(5 n - 5)}{[1 + a_{n - 1} a_{n}]}$
Hence, the correct option is (C).

Suggest Corrections

Similar questions

Q. If

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

are in A.P. and

a_{i} > 0

for all i, then 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}}

Q. If

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

are in A.P. and

a_{i} > 0

for all i, then 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}}

Q. If

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

are in A.P. where

a_{i} > 0

for all i, show that

\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . + \frac{1}{\sqrt{a_{n - 1} + \sqrt{a_{n}}}} = \frac{(n - 1)}{\sqrt{a_{1} + \sqrt{a_{n}}}}

If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =$

Join BYJU'S Learning Program

If a1,a2,a3,…. an are in AP with common difference 5 and if aiaj≠-1fori,j=1,2,….N, then tan-15[1+a1a2]+tan-15[1+a2a3]+….+tan-15[1+an-1an]=

If $a_{1}, a_{2}, a_{3}, \dots .$ an are in AP with common difference $5$ and if $a_{i} a_{j} \neq - 1 f o r i, j = 1, 2, \dots . N,$ then $\tan^{- 1} \frac{5}{[1 + a_{1} a_{2}]} + \tan^{- 1} \frac{5}{[1 + a_{2} a_{3}]} + \dots . + \tan^{- 1} \frac{5}{[1 + a_{n - 1} a_{n}]} =$