If the n^{th} term of an A.P. is (2n - 1),the | vedclass

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

Question

(C) Given that the $n^{th}$ term is $T_n = 2n - 1$.The first term $a = T_1 = 2(1) - 1 = 1$.The last term $l = T_n = 2n - 1$.The sum of the first $n$ t

Vedclass · Accepted Answer

$n^2$

Vedclass · Answer

(C) Given that the $n^{th}$ term is $T_n = 2n - 1$.The first term $a = T_1 = 2(1) - 1 = 1$.The last term $l = T_n = 2n - 1$.The sum of the first $n$ terms of an $A.P.$ is given by $S_n = \frac{n}{2}(a + l)$.Substituting the values,we get $S_n = \frac{n}{2}(1 + 2n - 1) = \frac{n}{2}(2n) = n^2$.Alternatively,$S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (2k - 1) = 2\sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 = 2 \cdot \frac{n(n+1)}{2} - n = n^2 + n - n = n^2$.

If the $n^{th}$ term of an $A.P.$ is $(2n - 1)$,then the sum of its first $n$ terms is

Similar Questions

If the sum of the first $11$ terms of an $A.P.$,$a_{1}, a_{2}, a_{3}, \ldots$ is $0$ $(a_{1} \neq 0)$,then the sum of the $A.P.$,$a_{1}, a_{3}, a_{5}, \ldots, a_{23}$ is $k a_{1}$,where $k$ is equal to

If $\frac{1}{p + q}, \frac{1}{r + p}, \frac{1}{q + r}$ are in $A.P.$,then

The maximum sum of the series $20 + 19\frac{1}{3} + 18\frac{2}{3} + \dots$ is

Find the $17^{\text{th}}$ and $24^{\text{th}}$ term in the following sequence whose $n^{\text{th}}$ term is $a_{n} = 4n - 3$.

The number of terms of the $A.P. 3, 7, 11, 15, ...$ to be taken so that the sum is $406$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module