If a_{1}, a_{2}, a_{3}, . . ., a_{n} are in AP with common difference 5 and if a_{i} a_{j} \neq-1 for i, j=1,2, . . ., n then \tan ^{-1}( {5}/{1+a_{1} a_{2}})+\tan ^{-1}( {5}/{1+a_{2} a_{3}}) +. . .+\tan ^{-1}( {5}/{1+a_{n-1} a_{n}}) is equal to

Correct Answer is : tan−1(5n−51+ana1)

\tan ^{-1}( {5}/{1+a_{n} a_{n-1}})

\tan ^{-1}( {5 a_{1}}/{1+a_{n} a_{1}})

\tan ^{-1}( {5 n-5}/{1+a_{n} a_{1}})

\tan ^{-1}( {5 n-5}/{1+a_{1} a_{n+1}})

\tan ^{-1}( {5 n}/{1+a_{1} a_{n}})

KEAM 2010 Math Paper

Question : 79 of 120

Marks: +1, -0

If a1,a2,a3,...,an are in AP with common difference 5 and if aiaj≠−1 for i,j=1,2,...,n then

tan−1(‌

1+a1a2

)+tan−1(‌

1+a2a3

)+...+tan−1(‌

1+an−1an

)

is equal to

tan−1(‌
5
1+anan−1
)
tan−1(‌
5a1
1+ana1
)
tan−1(‌
5n−5
1+ana1
)
tan−1(‌
5n−5
1+a1an+1
)
tan−1(‌
5n
1+a1an
)

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