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 :
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$n(100-n+1)=488 \times 5$
$n^{2}-101 n+488 \times 5=0$
$n=61,40$
$\mathrm{T}_{\mathrm{n}}=\mathrm{a}+(\mathrm{n}-1) \mathrm{d}=\frac{100}{5}-\frac{2}{5} \times 60$
$=20-24=-4$