If S=∑r=0991100 sin(πr200) and R=∑r=11001100 sin(πr200) , then

Let f ( x ) = sin ( π x 2 ) f(x) is increasing in [ 0 , 1 ] and ∫ 0 1 sin π x 2 d x = 2 π So, S < 2 π < R <img 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qYIIA5T6SJYCPhBAHH4kQeigIApAojDVLoIFgJ+EEAcfuSBKCBgigDiMJUugoWAHwQQhx95IAoImCKAOEyli2Ah4AcBxOFHHogCAqYIIA5T6SJYCPhBAHH4kQeigIApAojDVLoIFgJ+EEAcfuSBKCBgigDiMJUugoWAHwQQhx95IAoImCKAOEyli2Ah4AcBxOFHHogCAqYIIA5T6SJYCPhBAHH4kQeigIApAojDVLoIFgJ+EEAcfuSBKCBgigDiMJUugoWAHwQQhx95IAoImCKAOEyli2Ah4AcBxOFHHogCAqYIIA5T6SJYCPhBAHH4kQeigIApAojDVLoIFgJ+EEAcfuSBKCBgigDiMJUugoWAHwQQhx95IAoImCKAOEyli2Ah4AcBxOFHHogCAqYIIA5T6SJYCPhBAHH4kQeigIApAojDVLoIFgJ+EEAcfuSBKCBgigDiMJUugoWAHwQQhx95IAoImCKAOEyli2Ah4AcBxOFHHogCAqYIIA5T6SJYCPhBAHH4kQeigIApAojDVLoIFgJ+EEAcfuSBKCBgigDiMJUugoWAHwQQhx95IAoImCKAOEyli2Ah4AcBxOFHHogCAqYImBNHZWWl1NXVSVVVlSnQBAuBPBEwJ448wedaIGCVAOKwmjnihkCGBBBHhvBpGgJWCSAOq5kjbghkSABxZAifpiFglQDisJo54oZAhgQQR4bwaRoCVgn8D1OOwZCg6Ui6AAAAAElFTkSuQmCC">

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If $S = \sum_{r = 0}^{99} \frac{1}{100} \sin (\frac{π r}{200})$ and $R = \sum_{r = 1}^{100} \frac{1}{100} \sin (\frac{π r}{200})$ , then

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$R > \frac{2}{π}$

$S = R = \frac{2}{π}$

$S < R < \frac{2}{π}$

$S < \frac{2}{π}$

answer is B, C.

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Detailed Solution

Let $f (x) = \sin (\frac{π x}{2})$
f(x) is increasing in $[0, 1]$ and $\int_{0}^{1} \sin \frac{π x}{2} d x = \frac{2}{π}$
So, $S < \frac{2}{π} < R$
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