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,we substitute $n = 1, 2, 3, 4, 5$:For $n = 1: T_1 = 3(1) - 1

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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,we substitute $n = 1, 2, 3, 4, 5$:For $n = 1: T_1 = 3(1) - 1 = 2$For $n = 2: T_2 = 3(2) - 1 = 5$For $n = 3: T_3 = 3(3) - 1 = 8$For $n = 4: T_4 = 3(4) - 1 = 11$For $n = 5: T_5 = 3(5) - 1 = 14$The first five terms are $2, 5, 8, 11, 14$.The sum of the first five terms is $S_5 = 2 + 5 + 8 + 11 + 14 = 40$.Alternatively,using the sum formula: $S_n = \sum_{k=1}^{n} (3k - 1) = 3 \sum_{k=1}^{n} k - \sum_{k=1}^{n} 1 = 3 \frac{n(n+1)}{2} - n$.For $n = 5: S_5 = \frac{3(5)(6)}{2} - 5 = 45 - 5 = 40$.

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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