If the sum of the series $20 + 19\frac{3}{5} + 19\frac{1}{5}\, + \,18\frac{4}{5} + \,...\,\,{\rm{upto}}\,\,\,n{\rm{th}}\,$ term is 488 and the $n\,{\rm{th}}$ term is negative, then

$\,n{\rm{th}}\,\,{\rm{term}}\,{\rm{is}}\, - 4$

$n{\rm{th}}\,\,{\rm{term}}\,\,{\rm{is}}\, - 4\frac{2}{5}$

JEE MATHEMATICS MCQ: SEQUENCE OF SERIES

1 Easy

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Application +4 | -1

If the sum of the series

20 + 19\frac{3}{5} + 19\frac{1}{5}\, + \,18\frac{4}{5} + \,...\,\,\mathrm{upto}\,\,\,n\mathrm{th}\,

term is 488 and the

n\,\mathrm{th}

term is negative, then

View answer and explanation

Correct option: A

\,n\mathrm{th}\,\,\mathrm{term}\,\mathrm{is}\, - 4

Option A (

\n\mathrm{th}\,\mathrm{term}\,\mathrm{is}\, - 4

) is correct because the nth term is -4 when the sum of the series is 488 and the nth term is negative.

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