Question

Which is the first negative term of the AP $20,\frac{77}{4},\frac{37}{2},\frac{71}{4}$.

Answer 1

Determine the first negative term of the AP $20,\frac{77}{4},\frac{37}{2},\frac{71}{4}$.

By given series calculate the first term and its difference:

First-term $a=20$

$$\begin{gathered} Difference\left(d\right)=secondterm-firstterm \\ \Rightarrow =\frac{77}{4}-20 \\ \Rightarrow =-\frac{3}{4} \end{gathered}$$

Finding the first negative term of the AP series:

$T_n=0$

Using formula.

$$\begin{gathered} T_n=a+(n-1)d \\ \Rightarrow 0=20+(n-1)(-\frac{3}{4}) \\ \Rightarrow 0=20-\frac{3}{4}n+\frac{3}{4} \\ \Rightarrow \frac{3}{4}n=20+\frac{3}{4} \\ \Rightarrow \frac{3}{4}n=\frac{83}{4} \\ \Rightarrow n=\frac{83}{3} \\ \Rightarrow n=27.66 \end{gathered}$$

Hence, the ${28}^{th}$ term is the first negative term and its value can be defined as follows:
$$\begin{gathered} T_{28}=a+27d \\ =20+27(-\frac{3}{4}) \\ =20+(-\frac{81}{4}) \\ =\frac{(80-81)}{4} \\ =-\frac{1}{4} \end{gathered}$$
Hence, the $-\frac{1}{4}$ is the first negative term of the given Arithmetic Progression.