Question

How many terms of the $AP:20+19\frac{1}{3}+18\frac{2}{3}+.....$ must be taken so that their sum is $300$?

Answer 1

$\begin{array}{c}a=20d=19\frac{1}{3}-20=\frac{58-60}{3}=\frac{-2}{3}\\ S_n=300\\ \Rightarrow S_n=\frac{n}{2}\left(2a+\left(n-1\right)d\right)\\ \Rightarrow 300=\frac{n}{2}\left(2\times 20=\left(n-1\right)\left(\frac{-2}{3}\right)\right)\\ \Rightarrow 600=n\left(40-\frac{2n}{3}+\frac{2}{3}\right)\\ \Rightarrow 600=n\left(\frac{120-2n+2}{3}\right)\\ \Rightarrow 1800=122n-2n^2\\ \Rightarrow n^2-61n+900=0\\ \left(n-25\right)\left(n-36\right)=0\\ n=25,36\end{array}$