Question

. A sequence starts with 180, 120, …. If the sequence is arithmetic, create a recursive and explicit formula for the sequence. If the sequence is geometric, create a recursive and explicit formula for the sequence.

Answer 1

Answer:

Arithmetic Sequence

$T_n = 240-60n$ ---- Explicit

$T_n = T_{n-1} - 60$ --- Recursive

Geometric Sequence

$T_n = 270* (\frac{2}{3})^n$ ---- Explicit

$T_n = T_{n-1} * \frac{2}{3}$ ---- Recursive

Step-by-step explanation:

Given

$180, 120,....$

(a) Assume it is an arithmetic sequence

The explicit formula is calculated using:

$T_n = a + (n - 1)d$

Where

$a = 180$

$d = 120 - 180$

$d = -60$

So, we have:

$T_n = 180 + (n - 1)*-60$

$T_n = 180 -60n + 60$

Rewrite

$T_n = 180 + 60-60n$

$T_n = 240-60n$

The recursive function is calculated using:

$T_1 = 180$

$T_2 = 120 = 180 - 60 = T_1 - 60$

$T_3 = 60 = 120 - 60 = T_2 -60$

-

$T_n = T_{n-1} - 60$

(b) Assume it is a geometric sequence

The explicit formula is calculated using:

$T_n = ar^{n-1}$

Where

$a = 180$

$r = \frac{120}{180}$

$r = \frac{2}{3}$

So, we have:

$T_n = 180 * (\frac{2}{3})^{n-1}$

Split

$T_n = 180 * (\frac{2}{3})^n \div (\frac{2}{3})^1$

$T_n = 180 * (\frac{2}{3})^n \div (\frac{2}{3})$

Rewrite as:

$T_n = 180 * (\frac{2}{3})^n * (\frac{3}{2})$

$T_n = 180 * (\frac{3}{2})* (\frac{2}{3})^n$

$T_n = 180 * \frac{3}{2}* (\frac{2}{3})^n$

$T_n = 90 * 3* (\frac{2}{3})^n$

$T_n = 270* (\frac{2}{3})^n$

The recursive function is calculated using:

$T_1 = 180$

$T_2 = 120 = 180 * \frac{2}{3} = T_1 * \frac{2}{3}$

$T_3 = 80 = 120 * \frac{2}{3} = T_2 * \frac{2}{3}$

-

$T_n = T_{n-1} * \frac{2}{3}$