Question

The maximum sum of the series $20+19\frac{1}{3}+18\frac{2}{3}+...$ is

• A. $310$
• B. $300$
• C. $320$
• D. none of these

Answer 1

The correct option is A

$310$

Explanation for correct option

Given series, $20+19\frac{1}{3}+18\frac{2}{3}+...$ are in A.P.

Here, $d=-\frac{2}{3},a=\frac{60}{3}$

Since the A.P. is decreasing, the last term would be $0$. After $0$ the sum of the series will tend top decrease.

So for maximum value the last term should be $0$.

Now, to find the number of terms,

$$\begin{gathered} a_n=0 \\ \Rightarrow a+(n-1)d=0 \\ \Rightarrow 20-(n-1)\frac{2}{3}=0 \\ \Rightarrow (n-1)2=60 \\ \Rightarrow n-1=30 \\ \Rightarrow n=31 \end{gathered}$$

The Sum of the series is given as:

$S_n=\frac{n}{2}(\mathrm{first}\mathrm{term}+\mathrm{last}\mathrm{term})$

$\Rightarrow S_{31}=\left(\frac{31}{2}\right)\left(\frac{60}{3}\right)=310$

Hence, $\mathrm{Option}\left(\mathrm{A}\right)$ is the answer.