JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

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The minimum number of terms from the beginning of the series

20 + 22\frac{2}{3} + 25\frac{1}{3} + \ldots

so that the sum may exceed 1568, is

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Correct option: D

Option D is correct because the minimum number of terms is found by solving the inequality for the sum of an arithmetic series exceeding 1568.

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A=\{1,2,3\}, B=\{3,4\}, C=\{4,5,6\}

{A} \cup({B} \cap {C})

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(B \cup C)-A

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{x^2},{b^2},{y^2}

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\log \left(\frac{a}{3 b}\right)

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\frac{3}{4}