Question

If the sum of the series $20+19\frac{3}{5}+19\frac{1}{5}+18\frac{4}{5}+........$up to $nth$ term is $488$ and the $nth$ term is negative, then:

• A. $n=60$
• B. $n=41$
• C. $nth$ term is $-4$
• D. $nth$ term is $-4\frac{2}{5}$

Answer 1

The correct option is C

$nth$ term is $-4$

Explanation for the correct answer.

The given series $20+19\frac{3}{5}+19\frac{1}{5}+18\frac{4}{5}+........$is in A.P, where $a=20\text{and}d=-\frac{2}{5}$.

Sum of $n$ terms of A.P, ${\mathit{S}}_{\mathbf{n}}\mathbf{=}\frac{\mathbf{n}}{\mathbf{2}}\mathbf{[}\mathbf{2}\mathbf{a}\mathbf{+}\left(n-1\right)\mathbf{d}\mathbf{]}$.

$\begin{array}{rcl}\Rightarrow S_{\mathrm{n}}& =& \frac{\mathrm{n}}{2}[2\left(20\right)+\left(\mathrm{n}-1\right)\left(-\frac{2}{5}\right)]\\ \Rightarrow 488& =& \frac{\mathrm{n}}{2}\left[\left(\frac{202-2n}{5}\right)\right]\\ \Rightarrow 2440& =& 101n-n^2\\ & \Rightarrow & n^2-101n+2440=0\\ \Rightarrow \left(n-61\right)\left(n-40\right)& =& 0\\ \Rightarrow n& =& 61\mathrm{or}40\end{array}$

The $nth$term of A.P is $=T_n=a+\left(n-1\right)d$. So,

$$\begin{gathered} T_{61}=20+\left(60\times -\frac{2}{5}\right)=-4 \\ {T}_{40}=20+\left(39\times -\frac{2}{5}\right)=4.4 \end{gathered}$$

It is given that the $nth$ term is negative. Therefore, $n=61\text{and}{T}_n=-4$.

Hence, option C is correct.