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.