Can any term of the $A.P.$ $242, 236, 230, \ldots$ be $0$? If yes,which term?
(N/A) The given $A.P.$ is $242, 236, 230, \ldots$
Here,the first term $a = 242$ and the common difference $d = 236 - 242 = -6$.
Let the $n^{th}$ term of the $A.P.$ be $0$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values: $0 = 242 + (n - 1)(-6)$.
$-242 = -6(n - 1)$.
$n - 1 = \frac{242}{6} = \frac{121}{3} = 40.33$.
$n = 41.33$.
Since $n$ must be a positive integer,$0$ cannot be a term of this $A.P.$
Here,the first term $a = 242$ and the common difference $d = 236 - 242 = -6$.
Let the $n^{th}$ term of the $A.P.$ be $0$.
The formula for the $n^{th}$ term is $a_n = a + (n - 1)d$.
Substituting the values: $0 = 242 + (n - 1)(-6)$.
$-242 = -6(n - 1)$.
$n - 1 = \frac{242}{6} = \frac{121}{3} = 40.33$.
$n = 41.33$.
Since $n$ must be a positive integer,$0$ cannot be a term of this $A.P.$
Explore More
Similar Questions
The $21^{\text{st}}$ term of the $AP$ whose first two terms are $-3$ and $4$ is
Easy
View SolutionWhich term of the $AP : 121, 117, 113, \ldots$ is its first negative term?
Difficult
View SolutionDetermine whether the following sequence is an $A.P.$ or not. (Assume that the pattern continues.) If it is an $A.P.$,find its $n^{th}$ term: $1.4, 2.3, 3.2, 4.1, \dots$
Easy
View SolutionThe first term of a finite $A.P.$ is $5$ and its last term is $95$. If the common difference of the $A.P.$ is $5$,then there are $\ldots \ldots \ldots$ terms in the $A.P.$
Medium
View SolutionThe sum of $n$ terms of an $A.P.$ is given by $S_{n} = 5n^{2} - 3n$. Find the $n^{th}$ term of the $A.P.$
Medium
View SolutionVedclass Products
For Students
Vedclass Test Series
Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.
Start Free TrialFor Teachers
Exam Paper Generator
Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.
Try FreeFor Institutes
Online Exam Module
Live online exams with unlimited students, 360° analytics & white-label branding.
See Demo