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

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(C) Given that the $n^{th}$ term is $T_n = 2n - 1$.To find the sum of the first $n$ terms $(S_n)$,we can use the formula $S_n = \sum_{k=1}^{n} T_k$.$S

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${n^2}$

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(C) Given that the $n^{th}$ term is $T_n = 2n - 1$.To find the sum of the first $n$ terms $(S_n)$,we can use the formula $S_n = \sum_{k=1}^{n} T_k$.$S_n = \sum_{k=1}^{n} (2k - 1)$$S_n = 2 \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1$Using the sum of the first $n$ natural numbers formula $\sum_{k=1}^{n} k = \frac{n(n+1)}{2}$:$S_n = 2 \left[ \frac{n(n+1)}{2} \right] - n$$S_n = n(n + 1) - n$$S_n = n^2 + n - n = n^2$.Alternatively,the sequence is $1, 3, 5, \dots, (2n - 1)$,which is the sum of the first $n$ odd numbers,which is known to be $n^2$.

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

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