In=∫10xntan−1xdx. If anIn+2+bnIn=cn ∀ n

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I_{n} = \int_{0}^{1} x^{n} {tan}^{- 1} x d x . If a_{n} I_{n + 2} + b_{n} I_{n} = c_{n} \forall n ϵ N, n \geq 1

then

Options:

A . a1,a2,a3.....are in A.P.

B . b1,b2,b3.....are in G.P.

C . c1,c2,c3.....are in H.P.

D . a1,a2,a3.....are in H.P.

Answer: Option A
:
A

I_{n} = {(\frac{x^{n + 1}}{n + 1} t a n^{- 1} x)}_{0}^{- 1} - \int_{0}^{1} \frac{x^{n + 1}}{n + 1} . \frac{1}{1 + x^{2}} d x (n + 1) I_{n} = \frac{π}{4} - \int_{0}^{1} \frac{x^{n + 1}}{1 + x^{2}} d x (n + 3) I_{n + 2} = \frac{π}{4} - \int_{0}^{1} \frac{x^{n + 3}}{1 + x^{2}} d x ∴ (n + 1) I_{n} + (n + 3) I_{n + 2} = \frac{π}{2} - \frac{1}{n + 2} ∴ a_{n} = (n + 3) \Rightarrow a_{1}, a_{2}, a_{3}

.....are in A.P.

b_{n} = (n + 1) \Rightarrow b_{1}, b_{2}

..... are in A.P.

c_{n} = \frac{π}{2} - \frac{1}{n + 2}

not in any progression.

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