If θ=tan−1d1+a1a2+tan−1d1+a2a3+⋯+tan−1d1+an−1an, where a1,a2,a3,⋯an are in A.P. with common difference d, then tanθ=
A
(n−1)d1+a1an
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B
(n−1)da1+an
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C
an−a1an+a1
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D
nd1+a1an
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Solution
The correct option is A(n−1)d1+a1an θ=tan−1a2−a11+a1a2+tan−1a3−a21+a2a3+⋯+tan−1an−an−11+an−1an =(tan−1a2−tan−1a1)+(tan−1a3−tan−1a2)+⋯+tan−1an−an−11+an−1an+⋯+(tan−1an−tan−1an−1) =tan−1an−tan−1a1 =tan−1an−a11+a1an=tan−1(n−1)d1+a1an ∴tanθ=(n−1)d1+a1an