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

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(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

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$n^2$

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(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

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