A) A.P.
B) G.P.
C) H.P.
D) None of these
Correct Answer: A
Solution :
[a] \[y=\sin \pi \sqrt{x}\] meets x-axis when \[\pi \sqrt{x}=n\pi \]or \[x={{n}^{2}},\,\,n\in N\]. Therefore, the area of half-wave between \[x={{n}^{2}}\] and \[x={{(n+1)}^{2}}\] is \[{{S}_{n}}=\left| \int\limits_{{{n}^{2}}}^{{{(n+1)}^{2}}}{\sin \pi \sqrt{x}\,dx} \right|\] Putting \[\pi \sqrt{x}=y\] and therefore, \[{{\pi }^{2}}dx=2ydy,\] we get \[{{S}_{n}}=\left| \frac{2}{{{\pi }^{2}}}\int\limits_{n\pi }^{(n+1)\pi }{y\,\sin \,ydy} \right|\] \[=\left| \frac{2}{{{\pi }^{2}}}\left[ -(n+1)\pi cos(n+1)\pi +n\pi cosn\pi \right] \right|\] \[=\frac{2(2n+1)}{\pi },\,\,\,n\in N\] Thus, \[{{S}_{0}},{{S}_{1}},{{S}_{2}}....\].are in A.P.
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