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If $$a_n =\int^{\pi}_0 \frac{\sin(2n-1)x}{\sin x}dx$$ ,

then $$a_1,a_2,.....a_n$$ are in

(a) A.P and H.P

(b) A.P and G.P but not in H.P

(c) G.P and H.P

(d) A.P. ,G.P and H.P.

I have solved this problem , and getting all the terms equal to

$\pi$ i.e. $$a_1 =a_2=a_3 =.....\pi$$

But the answer given is option (b) however I think it should be (d) .....as in reasoning it is given that equal terms can be in A.P, GP but not in H.P ..request you to please clarify on this ...thanks....

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Consider $x\mapsto \dfrac x 2$ to get $$a_n =\frac 1 2\int^{\pi/2}_0 \frac{\sin\left(n-\frac 12\right)x}{\sin \dfrac x2}dx$$

Note then that $$a_{n+1} =\frac 1 2\int^{\pi/2}_0 \frac{\sin\left(n+\frac 12\right)x}{\sin \dfrac x2}dx$$

But $$\frac{\sin\left(n+\frac 12\right)x}{\sin \dfrac x2}= 1+2\sum_{k=1}^{n}\cos kx$$ Can you take it from here?

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  • $\begingroup$ I have mentioned I have solved this problem.. request you to please see my problem.. thanks for your answer any way... $\endgroup$
    – Sachin
    Aug 13, 2013 at 4:09

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