If twice the 11^{th} term of an Arithmetic Pr | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Given the formula for the $n^{th}$ term of an Arithmetic Progression: $t_n = a + (n - 1)d$.For the $11^{th}$ term: $t_{11} = a + 10d$.For the $21^

Vedclass · Accepted Answer

$0$

Vedclass · Answer

(C) Given the formula for the $n^{th}$ term of an Arithmetic Progression: $t_n = a + (n - 1)d$.For the $11^{th}$ term: $t_{11} = a + 10d$.For the $21^{st}$ term: $t_{21} = a + 20d$.According to the problem: $2t_{11} = 7t_{21}$.$2(a + 10d) = 7(a + 20d)$.$2a + 20d = 7a + 140d$.$5a + 120d = 0$.Dividing by $5$: $a + 24d = 0$.Now,the $25^{th}$ term is $t_{25} = a + (25 - 1)d = a + 24d$.Since $a + 24d = 0$,therefore $t_{25} = 0$.

If twice the $11^{th}$ term of an Arithmetic Progression is equal to seven times its $21^{st}$ term,then its $25^{th}$ term is .......

Similar Questions

If $1, \log _9(3^{1-x}+2), \log _3(4 \cdot 3^x-1)$ are in $A.P.$, then $x$ equals

If $a, b, c, d, e, f$ are in an arithmetic progression,then $e - c = \dots$

If the sum of $n$ terms of an $A.P.$ is $nA + n^2B$,where $A$ and $B$ are constants,then its common difference will be

Let $\sum_{k=1}^{n} a_{k} = \alpha n^{2} + \beta n$. If $a_{10} = 59$ and $a_{6} = 7a_{1}$, then $\alpha + \beta$ is equal to:

The sum of the first four terms of an $A.P.$ is $56$. The sum of the last four terms is $112$. If its first term is $11$,then find the number of terms.

Vedclass Test Series

Exam Paper Generator

Online Exam Module