The maximum sum of the series 20 + 19frac{1}{ | vedclass

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(A) The given series is an arithmetic progression with the first term $a = 20$ and common difference $d = 19\frac{1}{3} - 20 = -\frac{2}{3}$.The $n^{t

Vedclass · Accepted Answer

$310$

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(A) The given series is an arithmetic progression with the first term $a = 20$ and common difference $d = 19\frac{1}{3} - 20 = -\frac{2}{3}$.The $n^{th}$ term of the series is given by $a_n = a + (n - 1)d = 20 + (n - 1)\left( -\frac{2}{3} ight)$.For the sum to be maximum,we consider terms until they become non-negative,i.e.,$a_n \ge 0$.$20 - \frac{2}{3}(n - 1) \ge 0$$20 \ge \frac{2}{3}(n - 1)$$30 \ge n - 1$$n \le 31$.Thus,the sum of the first $31$ terms is the maximum sum.The sum $S_n = \frac{n}{2}[2a + (n - 1)d]$.$S_{31} = \frac{31}{2}[2(20) + (31 - 1)(-\frac{2}{3})]$$S_{31} = \frac{31}{2}[40 + 30(-\frac{2}{3})]$$S_{31} = \frac{31}{2}[40 - 20] = \frac{31}{2} 	imes 20 = 310$.

The maximum sum of the series $20 + 19\frac{1}{3} + 18\frac{2}{3} + \dots$ is

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