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B
1I2+I4,1I3+I5,1I4+I6are in A.P.
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C
I2+I4,I3+I5,I4+I6are in A.P.
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D
I2+I4,(I3+I5)2,I4+I6are in G.P.
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Solution
The correct option is B1I2+I4,1I3+I5,1I4+I6are in A.P. In+2+In=π2∫π4cotnx⋅cosec2xdx=[−(cotx)n+1n+1]π2π4 ⇒In+2+In=1n+1 ⇒I2+I4=13,I3+I5=14,I4+I6=15
So, 1I2+I4,1I3+I5,1I4+I6 are in A.P.