MathematicsEasy210×since 2002Q3433If bn=∫0π2cos2nxsinxdx, n∈N{b_n} = \int_0^{{\pi \over 2}} {{{{{\cos }^2}nx} \over {\sin x}}dx,\,n \in N}bn=∫02πsinxcos2nxdx,n∈N, thenAb3−b2, b4−b3, b5−b4{b_3} - {b_2},\,{b_4} - {b_3},\,{b_5} - {b_4}b3−b2,b4−b3,b5−b4 are in A.P. with common difference −-−2B1b3−b2,1b4−b3,1b5−b4{1 \over {{b_3} - {b_2}}},{1 \over {{b_4} - {b_3}}},{1 \over {{b_5} - {b_4}}}b3−b21,b4−b31,b5−b41 are in an A.P. with common difference 2Cb3−b2, b4−b3, b5−b4{b_3} - {b_2},\,{b_4} - {b_3},\,{b_5} - {b_4}b3−b2,b4−b3,b5−b4 are in a G.P.D1b3−b2,1b4−b3,1b5−b4{1 \over {{b_3} - {b_2}}},{1 \over {{b_4} - {b_3}}},{1 \over {{b_5} - {b_4}}}b3−b21,b4−b31,b5−b41 are in an A.P. with common difference −-−2Check answerSkip