Question

Which term of the $A.P.$, $84,80,76....$ is $0$?

Answer 1

Hint: In this question we have been given an arithmetic progression sequence for which we have to find the number of the term when the value of the term is $0$. We will solve this question by first writing the first term of the sequence and the common difference between the two terms in the arithmetic progression. We will then use the formula ${{a}_{n}}=a+\left( n-1 \right)d$, where $a$ is the first term, $n$ is the number of the terms and $d$ is the common difference. We will substitute all the values and solve for $n$ to get the required solution.

Complete step by step answer:
We have the sequence given to us as:
$\Rightarrow 84,80,76....$
We can see from the sequence that the first term is $84$ therefore, we can write:
$\Rightarrow a=84$
Now to find the common difference we do $80-84$, which gives us $-4$ therefore, we can write:
$\Rightarrow d=-4$
Now we know that we have a ${{n}^{th}}$ term in the $A.P$ which is $0$ therefore we have ${{a}_{n}}=0$.
On substituting the values in the formula, we get:
$\Rightarrow 0=84+\left( n-1 \right)\times -4$
On multiplying the terms, we get:
$\Rightarrow 0=84-4n+4$
On simplifying the terms, we get:
$\Rightarrow 0=88-4n$
On transferring $4n$ from the right-hand side to the left-hand side, we get:
$\Rightarrow 4n=88$
On rearranging the terms, we get:
$\Rightarrow n=\dfrac{88}{4}$
On simplifying the terms, we get:
$\Rightarrow n=22$, which is the required of $n$.
Therefore, in the $A.P.$, $84,80,76....$ the ${{22}^{nd}}$ term is $0$, which is the required solution.

Note: It is to be noted that in an arithmetic progression the common difference can be positive as well as a negative value. When the common difference is negative the $A.P$ is a decreasing $A.P$. When the common difference is positive the $A.P$ is an increasing $A.P$. There also exists $G.P$ which stands for geometric progression which is for a sequence with terms with common multiple rather than a difference.