Question

An arithmetic sequence has this recursive formula: (a^1 =8, a^n= a^n-1 -6

Answer 1

Answer:

$a_n = 8 + (n - 1) (-6)$

Step-by-step explanation:

Given

$a_1 = 8$

Recursive: $a_{n} = a_{n-1} - 6$

Required

Determine the formula

Substitute 2 for n to determine $a_2$

$a_{2} = a_{2-1} - 6$

$a_{2} = a_{1} - 6$

Substitute $a_1 = 8$

$a_2 = 8 - 6$

$a_2 = 2$

Next is to determine the common difference, d;

$d = a_2 - a_1$

$d = 2 - 8$

$d = -6$

The nth term of an arithmetic sequence is calculated as

$a_n = a_1 + (n - 1)d$

Substitute $a_1 = 8$ and $d = -6$

$a_n = a_1 + (n - 1)d$

$a_n = 8 + (n - 1) (-6)$

Hence, the nth term of the sequence can be calculated using$a_n = 8 + (n - 1) (-6)$