A person is to count 4500 currency notes. Let a_n denote the number of notes he counts in the n^th minute. If a_1 = a_2 = = a_10 = 150 and a_10, a_11, are in A.P. with common difference -2 , then the time taken by him to count all notes is:

A person is to count 4500 currency notes. Let a_n denote the number of notes he counts in the n^th minute. If a_1 = a_2 = = a_10 = 150 and a_10, a_11, - Mathematics MCQ AIEEE 2026 | ExamTest.live

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A person is to count 4500 currency notes. Let a_{n} denote the number of notes he counts in the n^{th} minute. If a_{1} = a_{2} = \dots = a_{10} = 150 and a_{10}, a_{11}, \dots are in A.P. with common difference - 2, then the time taken by him to count all notes is:

Notes counted in first 10 mins = 10 \times 150 = 1500 .Remaining notes = 4500 - 1500 = 3000 .After n = 10, the sequence is 150, 148, 146, \dots Let he takes n more minutes.Sum = \frac{n}{2} [2 (148) + (n - 1) (- 2)] = 3000 (Note: a_{11} = 150 - 2 = 148). n [148 - n + 1] = 3000 n (149 - n) = 3000 ⟹ n^{2} - 149 n + 3000 = 0 .Roots: n = \frac{149 \pm 14 9 ^{2} - 12000}{2} = \frac{149 \pm 22201 - 12000}{2} = \frac{149 \pm 101}{2} . n_{1} = 125, n_{2} = 24 . n = 125 is not possible as a_{n} becomes negative.So n = 24 .Total time = 10 + 24 = 34 minutes.

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A person is to count 4500 currency notes. Let a_{n} denote the number of notes he counts in the n^{th} minute. If a_{1} = a_{2} = \dots = a_{10} = 150 and a_{10}, a_{11}, \dots are in A.P. with common difference - 2, then the time taken by him to count all notes is:

34 minutes

125 minutes

135 minutes

24 minutes

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