Question

Which term of the A.P. $12,7,2,-3,....$ is $-98$ ?

Answer 1

Hint: First, we have to select the formula which is to be used here. So, here we have to find which term will be equal to $-98$ so, we will use the formula ${{T}_{n}}=a+\left( n-1 \right)d$ to find the value of n. Remaining values are given in the question.

Complete step-by-step answer:

In this question, we are given the series of Arithmetic progression i.e. $12,7,2,-3,....$ and we have to find in this series which term is $-98$ .
So, we will use the formula of A.P. for finding the ${{n}^{th}}$ in series which is given by ${{T}_{n}}=a+\left( n-1 \right)d$ where a is first term in the series, d is the common difference between any two consecutive number, n is the number of term we want to find and ${{T}_{n}}$ is term for which we want to find value of n.
So, here we have $a=12$ , $d={{T}_{2}}-{{T}_{1}}=7-15=-5$ , ${{T}_{n}}=-98$ . So, substituting all these values in the given formula i.e. ${{T}_{n}}=a+\left( n-1 \right)d$ so, we get as,
${{T}_{n}}=a+\left( n-1 \right)d$
$\Rightarrow -98=12+\left( n-1 \right)\left( -5 \right)$
On multiplying the brackets, we get
$\Rightarrow -98=12-5n+5$
Now, taking variable term one side and constant term on the other side, we get as:
$\Rightarrow 5n=12+5+98$
$\Rightarrow 5n=115$
$\Rightarrow n=\dfrac{115}{5}=23$
So, the value of n is 23.
Thus, ${{23}^{rd}}$ term in this A.P. series is $-98$ .

Note: To find whether the obtained answer is correct then we can do verification by placing the value of n in the formula of finding ${{n}^{th}}$ term. S, here we can do verification done as below:
We have here, $a=12$ , $n=23$ , $d=-5$ . So, keeping all the values in the formula ${{T}_{n}}=a+\left( n-1 \right)d$ ,we get as:
${{T}_{n}}=a+\left( n-1 \right)d$
${{T}_{n}}=12+\left( 23-1 \right)\left( -5 \right)$
${{T}_{n}}=12-115+5$
${{T}_{n}}=-98$ .
Thus, we got the term for which we got $n=23$ .So, its verified.
Also, sometimes there are chances students make mistakes in taking formula ${{S}_{n}}=\dfrac{n}{2}\left( a+l \right)$ instead of ${{T}_{n}}=a+\left( n-1 \right)d$ so, be careful while taking it.