(B) Given that,$a = 5, d = 3, a_{n} = 50$.Using the formula for the $n^{th}$ term of an $AP$:$a_{n} = a + (n - 1)d$$50 = 5 + (n - 1)3$$45 = (n - 1)3$$

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(B) Given that,$a = 5, d = 3, a_{n} = 50$.Using the formula for the $n^{th}$ term of an $AP$:$a_{n} = a + (n - 1)d$$50 = 5 + (n - 1)3$$45 = (n - 1)3$$15 = n - 1$$n = 16$.Now,using the sum formula $S_{n} = \frac{n}{2}[a + a_{n}]$:$S_{16} = \frac{16}{2}[5 + 50]$$S_{16} = 8 \times 55$$S_{16} = 440$.Thus,$n = 16$ and $S_{n} = 440$.