Question

Which term of the AP 20, $19\frac{1}{4}$, $18\frac{1}{2}$, $17\frac{3}{4}$, ... is its first negative term?

Answer 1

The given AP is 20, $19\frac{1}{4}$, $18\frac{1}{2}$, $17\frac{3}{4}$, ... .

Here, a = 20 and d = $19\frac{1}{4}-20=\frac{77}{4}-20=\frac{77-80}{4}=-\frac{3}{4}$

Let the n^th term of the given AP be the first negative term. Then,

$$\begin{gathered} a_n<0 \\ \Rightarrow 20+\left(n-1\right)\times \left(-\frac{3}{4}\right)<0\left[a_n=a+\left(n-1\right)d\right] \\ \Rightarrow 20+\frac{3}{4}-\frac{3}{4}n<0 \\ \Rightarrow \frac{83}{4}-\frac{3}{4}n<0 \end{gathered}$$
$$\begin{gathered} \Rightarrow -\frac{3}{4}n<-\frac{83}{4} \\ \Rightarrow n>\frac{83}{3}=27\frac{2}{3} \\ \therefore n=28 \end{gathered}$$

Hence, the 28th term is the first negative term of the given AP.