If the sum of the series 20 + 19 frac{3}{5} + | vedclass

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(C) The given series is an Arithmetic Progression $(AP)$ with first term $a = 20$ and common difference $d = 19 \frac{3}{5} - 20 = \frac{98}{5} - \fra

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$n^{	ext{th}}$ term is $-4$

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(C) The given series is an Arithmetic Progression $(AP)$ with first term $a = 20$ and common difference $d = 19 \frac{3}{5} - 20 = \frac{98}{5} - \frac{100}{5} = -\frac{2}{5}$.The sum of $n$ terms is given by $S_n = \frac{n}{2} [2a + (n - 1)d] = 488$.Substituting the values: $\frac{n}{2} [2(20) + (n - 1)(-\frac{2}{5})] = 488$.$\frac{n}{2} [40 - \frac{2n}{5} + \frac{2}{5}] = 488$.$n [20 - \frac{n}{5} + \frac{1}{5}] = 488$.$n [\frac{100 - n + 1}{5}] = 488$.$n(101 - n) = 2440$.$n^2 - 101n + 2440 = 0$.Solving the quadratic equation: $(n - 40)(n - 61) = 0$.So,$n = 40$ or $n = 61$.If $n = 61$,the $n^{	ext{th}}$ term $T_n = a + (n - 1)d = 20 + (60)(-\frac{2}{5}) = 20 - 24 = -4$.If $n = 40$,the $n^{	ext{th}}$ term $T_n = 20 + (39)(-\frac{2}{5}) = 20 - 15.6 = 4.4$ (which is not negative).Since the $n^{	ext{th}}$ term must be negative,we take $n = 61$,and the $n^{	ext{th}}$ term is $-4$.

If the sum of the series $20 + 19 \frac{3}{5} + 19 \frac{1}{5} + 18 \frac{4}{5} + \ldots$ up to $n^{\text{th}}$ term is $488$ and the $n^{\text{th}}$ term is negative,then:

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