Mathematics
If the sum of the series
\(20 + 19\frac{3}{5} + 19\frac{1}{5} + 18\frac{4}{5} + \ldots\) up to
\(n^{\text{th}\text{~}}\) term is 488 and the
\(n^{\text{th}\text{~}}\) term is negative, then
A
\(n = 60\)
B
\(n^{\text{th}\text{~}}\) term is -4
C
\(n = 41\)
D
\(n^{\text{th}\text{~}}\) term is \(- 4\frac{2}{5}\)\\
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