The n^{th} term of an A.P. is 3n - 1. Choose | vedclass

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(D) Given the $n^{th}$ term of the $A.P.$ is $T_n = 3n - 1$.To find the first five terms,substitute $n = 1, 2, 3, 4, 5$:$T_1 = 3(1) - 1 = 2$$T_2 = 3(2

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$40$

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(D) Given the $n^{th}$ term of the $A.P.$ is $T_n = 3n - 1$.To find the first five terms,substitute $n = 1, 2, 3, 4, 5$:$T_1 = 3(1) - 1 = 2$$T_2 = 3(2) - 1 = 5$$T_3 = 3(3) - 1 = 8$$T_4 = 3(4) - 1 = 11$$T_5 = 3(5) - 1 = 14$The sum of the first five terms is $S_5 = 2 + 5 + 8 + 11 + 14 = 40$.Alternatively,using the formula for the sum of $n$ terms:$S_n = \sum_{k=1}^{n} (3k - 1) = 3 \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 = 3 \frac{n(n+1)}{2} - n$For $n = 5$:$S_5 = \frac{3 	imes 5 	imes 6}{2} - 5 = 45 - 5 = 40$.

The $n^{th}$ term of an $A.P.$ is $3n - 1$. Choose from the following the sum of its first five terms.

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