First slide
Harmonic Progression
Question

Let In=0π/4tannxdx. Then I2+I4,I3+I5,I4+I6,I5+I7,are in

Moderate
Solution

We have for r2,
Ir+Ir+2=0π/4tanrx1+tan2xdx              =0π/4tanrxsec2xdx             =1r+1tanr+1x0π/4=1r+1
Thus, the given sequence becomes,
                        13,14,15,16,

This is clearly an H.P.

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