In an A.P., given a = 5, d = 3, a_n = 50, find n and S_n.

By formula,

a_n = a + (n - 1)d

⇒ 50 = 5 + 3(n - 1)

⇒ 50 - 5 = 3(n - 1)

⇒ 45 = 3(n - 1)

⇒ n - 1 = 15

⇒ n = 1 + 15 = 16.

By formula,

S_n = $\dfrac{n}{2}[2a + (n - 1)d]$

Substituting values we get :

$\Rightarrow S_{16} = \dfrac{16}{2}[2 \times 5 + (16 - 1) \times 3] \\[1em] = 8 \times [10 + 15 \times 3] \\[1em] = 8 \times [10 + 45] \\[1em] = 8 \times 55 \\[1em] = 440.$

Hence, n = 16 and S_n = 440.