If a1, a2, a3 ...an are in AP with common difference 5 and if aiaj ≠ -1 then tan-1 5/(1+a1a2) + tan-1 5/(1+a2a3) +......+ tan-1 5/(1+an-1an) is equal to

(1) tan-1 5/(1+anan-1)

(2) tan-1 5a1/(1+ana1)

(3) tan-1 (5n-5)/(1+ana1)

(4) none of these

Solution:

Given a1, a2, a3 …an are in AP with d = 5

a2-a1 = a3-a2 = a4-a3 = ….= an-an-1 = 5

tan-1 5/(1+a1a2) + tan-1 5/(1+a2a3) +……+ tan-1 5/(1+an-1an) = [tan-1 (a2-a1)/(1+a1a2)] + [tan-1 (a3-a2)/(1+a3a2 )]+ …+[ tan-1 (an-an-1)/(1+anan-1)]

= tan-1 a2– tan-1 a1 + tan-1 a3– tan-1 a2+ .. tan-1 an– tan-1 an-1

= tan-1 an– tan-1 a1

= tan-1 (an-a1)/(1+a1an)

= tan-1 5(n-1)/(1+a1an)

= tan-1 (5n-5)/(1+a1an)

Hence option (3) is the answer.

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