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Mathematics Question on Differential equations

If bn=∫0π2cos2nxsinxdx,n∈N,b_n= ∫_0^{\frac{π}{2}} \frac{cos^2nx}{sinx }dx, n∈N, Then

A

b 3 – b 2, b 4 – b 3, b 5 – b 4 are in an A.P. with a common difference –2

B

1b3−b2,1b4−b3,1b5−b4, are in an A.P. with common difference 2\frac{1}{b_3-b_2},\frac{1}{b_4-b_3},\frac{1}{b_5-b_4}, \text{ are in an A.P. with common difference 2}

C

b 3 – b 2, b 4 – b 3, b 5 – b 4 are in a G.P.

D

1b3−b2,1b4−b3,1b5−b4, are in an A.P. with common difference -2\frac{1}{b_3-b_2},\frac{1}{b_4-b_3},\frac{1}{b_5-b_4}, \text{ are in an A.P. with common difference -2}

Answer

b 3 – b 2, b 4 – b 3, b 5 – b 4 are in an A.P. with a common difference –2

Explanation

Solution

The correct option is(D): 1b3−b2,1b4−b3,1b5−b4, are in an A.P. with common difference -2\frac{1}{b_3-b_2},\frac{1}{b_4-b_3},\frac{1}{b_5-b_4}, \text{ are in an A.P. with common difference -2}

If bn= ∫_0^{π/2} cos^2nx/sinx dx, n∈N, Then

are in A. P. with common difference –2.