Can any term of the $A.P.$ $242, 236, 230, \ldots$ be $0$? If yes,which term?

Vedclass pdf generator app on play store
Vedclass iOS app on app store
(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 $A.P.$

Explore More

Similar Questions

Find the number of terms in the finite $A.P.$ $6, 5.5, 5, \ldots, -12$.

The sum of the first twenty even natural numbers is $\ldots \ldots \ldots \ldots$

Difficult
View Solution

The sum of the first $30$ positive multiples of $6$ is:

Difficult
View Solution

The sequence $1, 3, 6, 10, \ldots$ is

For the finite $A.P.$ $3, 8, 13, \ldots, 253,$ find the $20^{th}$ term from the end.

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial
For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free
For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo