If a 1- a 2- - a n -1 are in A-P- with common difference d- then - r -1 n tan -1 d -1- a r a r -1-A- tan -1n-1 d-1-a1 an-1B- tan -1n-1 d-1-a1 an-1C- tan -1n-1 d-1-a1 an-1D- tan -1n d-1-a1 an-1

The correct option is D r = 1n∑ tan^ 1 ( d/1 + a_ra_r+1 ) = r = 1n∑ tan^ 1a_r+1 a_r/1 + a_r a_r+1 = r = 1n∑ (tan^ 1 a_r+1 tan^ 1a_r) = tan^ 1a_n+1 tan^ 1 a ...

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Byju's Answer

Standard XII

Mathematics

Algebra of Derivatives

If a 1, a 2, ...

Question

If $a_{1}, a_{2}, \dots \dots a_{n + 1}$ are in A.P. with common difference d, then
$n \sum r = 1 t a n^{- 1} (\frac{d}{1 + a_{r} a_{r + 1}}) =$

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Solution

The correct option is A
$n \sum r = 1 t a n^{- 1} (\frac{d}{1 + a_{r} a_{r + 1}}) = n \sum r = 1 t a n^{- 1} \frac{a_{r + 1} - a_{r}}{1 + a_{r} a_{r + 1}} = n \sum r = 1 (t a n^{- 1} a_{r + 1} - t a n^{- 1} a_{r}) = t a n^{- 1} a_{n + 1} - t a n^{- 1} a_{1} = t a n^{- 1} (\frac{a_{n + 1} - a_{r}}{1 + a_{r} a_{n + 1}}) = t a n^{- 1} (\frac{n d}{1 + a_{1} a_{n + 1}})$

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Similar questions

Q. If

a_{1}, a_{2}, a_{3}

,______ are in A.P. of common difference d, then prove that

S = {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} a_{n + 1}} = {tan}^{- 1} \frac{n d}{1 + a_{1} a_{n + 1}}

Q. If

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

is an A.P. with common difference d, then

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}}

.
Find the value of

p + q

Q. If

θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}

, where

a_{1}, a_{2}, a_{3}, \dots a_{n}

are in A.P. with common difference d, then

t a n θ =

Q.
lf

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

are in A.P. with common difference d, then

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

Q. If

θ = t a n^{- 1} \frac{d}{1 + a_{1} a_{2}} + t a n^{- 1} \frac{d}{1 + a_{2} a_{3}} + \dots + t a n^{- 1} \frac{d}{1 + a_{n - 1} a_{n}}

, where

a_{1}, a_{2}, a_{3}, \dots a_{n}

are in A.P. with common difference d, then

t a n θ =

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If a1,a2,……an+1 are in A.P. with common difference d, then n∑r=1tan−1(d1+arar+1)=

The correct option is A n∑r=1tan−1(d1+arar+1)=n∑r=1tan−1ar+1−ar1+arar+1=n∑r=1(tan−1ar+1−tan−1ar)=tan−1an+1−tan−1a1=tan−1(an+1−ar1+aran+1)=tan−1(nd1+a1an+1)

If $a_{1}, a_{2}, \dots \dots a_{n + 1}$ are in A.P. with common difference d, then
$n \sum r = 1 t a n^{- 1} (\frac{d}{1 + a_{r} a_{r + 1}}) =$