For any $A.P.$,$T_{30} - T_{20} = \ldots \ldots \ldots \ldots$
- A$10a$
- B$10d$
- C$T_{10}$
- D$10n$
Explore More
Similar Questions
Which of the following matchings is true for Part $I$ and Part $II$?
Part $I$ Part $II$ $1$. Sixth term of the $A.P.$ $1, 3, 5, 7, \ldots$ $a$. $105$ $2$. Eleventh term of the $A.P.$ $3, 6, 9, 12, \ldots$ $b$. $11$ $3$. Sixteenth term of the $A.P.$ $4, 6, 8, 10, \ldots$ $c$. $33$ $4$. Twenty-first term of the $A.P.$ $5, 10, 15, 20, \ldots$ $d$. $34$
| Part $I$ | Part $II$ |
| $1$. Sixth term of the $A.P.$ $1, 3, 5, 7, \ldots$ | $a$. $105$ |
| $2$. Eleventh term of the $A.P.$ $3, 6, 9, 12, \ldots$ | $b$. $11$ |
| $3$. Sixteenth term of the $A.P.$ $4, 6, 8, 10, \ldots$ | $c$. $33$ |
| $4$. Twenty-first term of the $A.P.$ $5, 10, 15, 20, \ldots$ | $d$. $34$ |
Difficult
View SolutionWhat is the common difference of an $AP$ in which $a_{18}-a_{14}=32$?
Easy
View SolutionVerify that each of the following is an $AP$,and then write its next three terms.
$a+b, (a+1)+b, (a+1)+(b+1), \ldots$
$a+b, (a+1)+b, (a+1)+(b+1), \ldots$
Medium
View SolutionJustify whether it is true to say that $-1, -\frac{3}{2}, -2, \frac{5}{2}, \ldots$ forms an $AP$ as $a_{2}-a_{1} = a_{3}-a_{2}$.
Medium
View SolutionThe sum of the first $30$ positive multiples of $6$ is:
Difficult
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