Question

If $I_n={\int }_0^{\pi /4}ta{n}^{n}xdx$, then $I_2+I_4,I_3+I_5,I_4+I_6,I_5+I_7$,....... are in

• A. A.P
• B. G.P
• C. H.P
• D. none of these

Answer 1

Answer: (C)

H.P

$I_n+I_{n+2}=\int (ta{n}^{n}x+ta{n}^{n+2}x)dx$

$I_n+I_{n+2}=\int ta{n}^{n}x.(1+ta{n}^{2}x)dx$

$I_n+I_{n+2}=\int ta{n}^{n}x.se{c}^{2}x.dx$

By applying the limits of integration, $I_n+I_{n+2}=\frac{ta{n}^{n+1}\frac{\pi }{4}}{n+1}=\frac{1}{n+1}$

Terms of the form $n+1$ form an AP for successive $n$ values.

Hence, their reciprocals $\frac{1}{n+1}$form an HP, for successive $n$ values.