If $T_{n} = 3 n - 1$ of an A.P., then

a = 2

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d = 3

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T_{5} = 15

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S_{3} = 15

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Solution

The correct option is D $S_{3} = 15$
Given : $T_{n} = 3 n - 1$
Now,
$a = T_{1} = 3 \times 1 - 1 = 2 \dots (i) a + d = T_{2} = 3 \times 2 - 1 = 5 \dots (i i) a + 2 d = T_{3} = 3 \times 3 - 1 = 8 \dots (i i i) a + 3 d = T_{4} = 3 \times 4 - 1 = 11 \dots (i v)$
Therefore, $a = 2, d = 3$
Now,
$S_{3} = T_{1} + T_{2} + T_{3} ∴ S_{3} = 2 + 5 + 8 = 15$

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