Sum of all the values of x in the interval [0, 314] satisfying the equation sin x = 0 is
A
4950π
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B
5050π
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C
5151π
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D
none of these
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Solution
The correct option is B 5050π sinx=0 x=nπ+0, where x∈[0,314] Thus, x=0,π,2π,3π,....100π Sum of all the values= 0+π+2π+...+100π = π(1+2+3+...+100) = π100×1012Using formula of an AP, sum=n(2a+(n−1)d)2 = 5050π