In the sum of the series

Question:

In the sum of the series

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

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

 

  1. (1) $n=60$

  2. (2) $n^{\text {th }}$ term is $-4$

  3. (3) $n=41$

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


Correct Option: , 2

Solution:

$S_{n}=20+19 \frac{3}{5}+19 \frac{1}{5}+18 \frac{4}{5}+\ldots$

$\because S_{n}=488$

$488=\frac{n}{2}\left[2\left(\frac{100}{5}\right)+(n-1)\left(-\frac{2}{5}\right)\right]$

$488=\frac{n}{2}(101-n) \Rightarrow n^{2}-101 n+2440=0$

$\Rightarrow n=61$ or 40

For $n=40 \Rightarrow T_{n}>0$

For $n=61 \Rightarrow T_{n}<0$

$n^{\text {th }}$ term $=T_{61}=\frac{100}{5}+(61-1)\left(-\frac{2}{5}\right)=-4$

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