AssertionA:sinπn+sin3πn+sin5πn+......+to n terms=0 Reason (R): sinα+sin(α+β)+........upto n terms=sinnβ2sinβ2sin2α+n-1β2

sin⁡πn+sin⁡3πn+sin⁡5πn+… to n terms =sin⁡n2.π2.nsin⁡2.π2.n⋅sin⁡2⋅πn+(n−1)⋅2πn2 Using, sin⁡α+sin⁡(α+β)+…+sin⁡(α+(n−1)β)=sin⁡nβ/2sin⁡β/2⋅sin⁡2α+(n−1)β2 =sin⁡πsin⁡π/nsin⁡2π+2nπ−2π2n=sin2⁡πsin⁡π/n=0∴ sin⁡πn+sin⁡3πn+sin⁡5πn+… to n terms =0A is true ,R is true .R⇒A

AssertionA:sinπn+sin3πn+sin5πn+......+to n terms=0 Reason (R): sinα+sin(α+β)+........upto n terms=sinnβ2sinβ2sin2α+n-1β2

sin⁡πn+sin⁡3πn+sin⁡5πn+… to n terms =sin⁡n2.π2.nsin⁡2.π2.n⋅sin⁡2⋅πn+(n−1)⋅2πn2 Using, sin⁡α+sin⁡(α+β)+…+sin⁡(α+(n−1)β)=sin⁡nβ/2sin⁡β/2⋅sin⁡2α+(n−1)β2 =sin⁡πsin⁡π/nsin⁡2π+2nπ−2π2n=sin2⁡πsin⁡π/n=0∴ sin⁡πn+sin⁡3πn+sin⁡5πn+… to n terms =0A is true ,R is true .R⇒A

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$Assertion (A) : \sin \frac{π}{n} + \sin \frac{3 π}{n} + \sin \frac{5 π}{n} + . . . . . . + t o n t e r m s = 0$
$R e a s o n (R) : \sin α + \sin (α + β) + . . . . . . . . u p t o n t e r m s = \frac{\sin \frac{n β}{2}}{\sin \frac{β}{2}} \sin (\frac{2 α + (n - 1) β}{2})$

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$A i s t r u e, R i s t r u e . R \Rightarrow A$

$A i s t r u e, R i s t r u e b u t R d o e s n o t i m p l y A$

$A is true, R is false$

$A is false, R is true$

answer is A.

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Detailed Solution

$\begin{array}{l} \sin \frac{π}{n} + \sin \frac{3 π}{n} + \sin \frac{5 π}{n} + \dots to n terms \\ = \frac{\sin \frac{n 2 . π}{2 . n}}{\sin \frac{2 . π}{2 . n}} \cdot \sin \{\frac{2 \cdot \frac{π}{n} + (n - 1) \cdot \frac{2 π}{n}}{2}\} \\ Using, \sin α + \sin (α + β) + \dots + \sin (α + (n - 1) β) = \frac{\sin n β / 2}{\sin β / 2} \cdot \sin \{\frac{2 α + (n - 1) β}{2}\} \\ = \frac{\sin π}{\sin π / n} \sin \{\frac{2 π + 2 n π - 2 π}{2 n}\} \\ = \frac{\sin^{2} π}{\sin π / n} = 0 \\ ∴ \sin \frac{π}{n} + \sin \frac{3 π}{n} + \sin \frac{5 π}{n} + \dots to n terms = 0 \end{array}$