Question

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

Answer 1

Hint: In this question, we are given an arithmetic progression series and we have to find its first negative term. For this, we will first determine the first term and common difference of this AP. Then we will suppose ${{n}^{th}}$ term to be the required negative term. Since, first negative term is required, so we will let ${{a}_{n}}<0$ where, ${{a}_{n}}$ represent ${{n}^{th}}$ term.
Using ${{a}_{n}}=a+\left( n-1 \right)d$ where a is first term and d is a common difference of AP, we will find the value of n and the required term will be the just greater whole number.

Complete step by step answer:
Here, we are given the arithmetic progression as:
121, 117, 113 . . . . . . . . . .
We have to find its first negative term. For this, let us first determine its first term and common difference.
As we can see, first term a = 121.
Common difference can be calculated as 117-121 = 113-117 = -4
Therefore, d = -4.
Now, let ${{n}^{th}}$ term of the AP is its first negative term. Therefore, ${{a}_{n}}$ will be negative and hence less than 0. Hence, ${{a}_{n}}<0$.
As we know, in an AP ${{n}^{th}}$ term can be calculated using formula ${{a}_{m}}=a+\left( m-1 \right)d$ where a is first term and d is common difference. So, let us use this formula for ${{a}_{n}}$.
$\begin{align}
  & {{a}_{n}}<0 \\
 & a+\left( n-1 \right)d<0 \\
\end{align}$
Putting value of a and d, we get:
\[\begin{align}
  & 121+\left( n-1 \right)\left( -4 \right)\text{ }<\text{ }0 \\
 & \Rightarrow 121-4\left( n-1 \right)\text{ }<\text{ }0 \\
 & \Rightarrow 121-4n+4\text{ }<\text{ }0 \\
 & \Rightarrow 125-4n\text{ }<\text{ }0 \\
 & \Rightarrow 125\text{ }<\text{ }4n \\
 & \Rightarrow \dfrac{125}{4}\text{ }<\text{ }n \\
\end{align}\]
As we can see, n is greater than \[\dfrac{125}{4}=31.25\].
Hence, we will take the whole number just after 31.25.
Therefore, the required term is ${{32}^{nd}}$ term.

Hence, ${{32}^{nd}}$ term will be the first negative term of the AP 121, 117, 113 . . . . . .

Note: Students should note that, while calculating common differences, make sure to subtract second term from first term. They should not get confused if the second term is smaller as the common difference can be negative too. For reaching from positive terms to negative terms, common differences need to be negative. Don't forget that, we have to take the whole number as required value, since there is never ${{31.25}^{th}}$ term. Also, the term taken should be just greater than the obtained value of n.