Question

Which term of the AP: $121,117,113,...$ is its first negative term?

Answer 1

Determine the first negative term.

Use the $n^{\mathrm{th}}$ term formula: $a_n=a+\left(n-1\right)d$

The given AP is $121,117,113,...$.

The first term is $a=121$.

The common difference is:

$\begin{array}{rcl}d& =& 117-121\\ & \Rightarrow & d=-4\end{array}$

Now,

$$\begin{gathered} a_n=a+\left(n-1\right)d \\ \Rightarrow a_n=121+\left(n-1\right)(-4) \\ \Rightarrow a_n=121-4n+4 \\ \Rightarrow a_n=125-4n \end{gathered}$$

Let $a_n$ be the first negative term of the given AP so

$$\begin{gathered} a_n<0 \\ \Rightarrow 125-4n<0 \\ \Rightarrow 125<4n \\ \Rightarrow n>\frac{125}{4} \\ \therefore n>31.25 \end{gathered}$$

So we can take $n=32$

Hence, the ${32}^{nd}$ term will be the first negative term of this AP.