If the sum of the series
Question:

If the sum of the series

$20+19 \frac{3}{5}+19 \frac{1}{5}+18 \frac{4}{5}+\ldots .$ upto $\mathrm{n}^{\text {th }}$ term is 488

and the $\mathrm{n}^{\text {th }}$ term is negative, then :

1. $\mathrm{n}^{\text {th }}$ term is $-4 \frac{2}{5}$

2. $\mathrm{n}=41$

3. $n^{\text {th }}$ term is $-4$

4. $\mathrm{n}=60$

Correct Option: , 3

Solution:

$\mathrm{S}=\frac{100}{5}+\frac{98}{5}+\frac{96}{5}+\frac{94}{5}+\ldots . . \mathrm{n}$

$\mathrm{S}_{\mathrm{n}}=\frac{\mathrm{n}}{2}\left(2 \times \frac{100}{5}+(\mathrm{n}-1)\left(-\frac{2}{5}\right)\right)=188$

v$n(100-n+1)=488 \times 5$

$n^{2}-101 n+488 \times 5=0$

$n=61,40$

$\mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}=\frac{100}{5}-\frac{2}{5} \times 60$

$=20-24=-4$