Question

The sum of the first n terms of an arithmetic sequence is n/2(4n + 20).

Answer 1

Answer:

(a).

$S_{n} = \frac{n}{2} (4n + 20) \\ \\S _{n - 1} = \frac{(n - 1)}{2} (4n - 4 + 20) \\ \\ S _{n - 1} = \frac{(n - 1)}{2} (4n + 16) \\ \\S _{n - 1} = \frac{(n - 1)(4n + 16)}{2} \\ \\ { \boxed{S _{n - 1} = {2 {n}^{2} + 6n - 8}}}$

(b).

from general equation:

$S _{n - 1} = \frac{(n - 1)}{2} (4n + 16)$

first term is 4n

common difference:

$16 = \{(n - 1) - 1 \}d \\ 16 = (n - 2)d \\ d = \frac{16}{n - 2}$