Find the maximum sum of the series 20 + 19fra | vedclass

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(A) The given series is an Arithmetic Progression $(AP)$ with the first term $a = 20$ and common difference $d = 19\frac{1}{3} - 20 = \frac{58}{3} - \

Vedclass · Accepted Answer

$310$

Vedclass · Answer

(A) The given series is an Arithmetic Progression $(AP)$ with the first term $a = 20$ and common difference $d = 19\frac{1}{3} - 20 = \frac{58}{3} - \frac{60}{3} = -\frac{2}{3}$.Since the common difference is negative,the terms will eventually become negative.The sum is maximum when we add only the positive terms.Let the $n$-th term be $t_n = a + (n-1)d$.We require $t_n \geq 0$,so $20 + (n-1)(-\frac{2}{3}) \geq 0$.$20 \geq \frac{2}{3}(n-1) \implies 30 \geq n-1 \implies n \leq 31$.Thus,the first $31$ terms are non-negative.The maximum sum is $S_{31} = \frac{31}{2} [2a + (31-1)d]$.$S_{31} = \frac{31}{2} [2(20) + 30(-\frac{2}{3})] = \frac{31}{2} [40 - 20] = \frac{31}{2} 	imes 20 = 310$.

Find the maximum sum of the series $20 + 19\frac{1}{3} + 18\frac{2}{3} + 18 + .....$

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